Question 12 Marks
Prove The Theorem : If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Answer
View full question & answer→Self
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Question 22 Marks
Prove The Theorem : The ratio of the intercepts made on a transversal by three parallel lines is equal to the ratio of the corrosponding intercepts made on any other transversal by the same parallel lines.
Answer
View full question & answer→Given : line $l \|$ line $m \|$ line $n$ $t_1$ and $t_2$ are transversals. Transversal $t_1$ intersects the lines in points A, B, C and $\mathrm{t}_2$ intersects the lines in points $\mathrm{P}$, Q, R. To prove : $\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{PQ}}{\mathrm{QR}}$
Question 32 Marks
In figure 1.73, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then $\frac{ A (\triangle ABC )}{ A (\triangle DCB )}=?$
Answer
View full question & answer→$\text { We know that, Area of triangle }=\frac{1}{2} \times \text { base } \times \text { height }$
$ \Rightarrow \frac{ A (\triangle ABC )}{ A (\triangle DCB )}=\frac{\frac{1}{2} \times BC \times AB }{\frac{1}{2} \times BC \times DC }$.
$ =\frac{ AB }{ DC }$
$=\frac{6}{8}$
$=\frac{3}{4}$
$ \Rightarrow \frac{ A (\triangle ABC )}{ A (\triangle DCB )}=\frac{\frac{1}{2} \times BC \times AB }{\frac{1}{2} \times BC \times DC }$.
$ =\frac{ AB }{ DC }$
$=\frac{6}{8}$
$=\frac{3}{4}$
Question 42 Marks
Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm then what is the corresponding base of the bigger triangle?
Answer
View full question & answer→(PROPERTY:Areas of triangles with equal heights are proportional to their corresponding bases.)
$\Rightarrow \frac{ A (\text { smaller triangle })}{ A (\text { bigger triangle })}=\frac{\text { base(smaller triangle })}{\text { base(bigger triangle) }} $
$ \Rightarrow \frac{2}{3}=\frac{6}{\text { base(bigger triangle) }}$
⇒ base(bigger triangle) = 9 cm
$\Rightarrow \frac{ A (\text { smaller triangle })}{ A (\text { bigger triangle })}=\frac{\text { base(smaller triangle })}{\text { base(bigger triangle) }} $
$ \Rightarrow \frac{2}{3}=\frac{6}{\text { base(bigger triangle) }}$
⇒ base(bigger triangle) = 9 cm
Question 52 Marks
Areas of two similar triangles are 225 sq.cm & 81 sq.cm. If a side of the smaller triangle is 12 cm, then find corresponding side of the bigger triangle.
Answer
View full question & answer→Let area of one(bigger) triangle be ' $A$ ', other(smaller) triangle be ' $B$ ',corresponding side of smaller triangle be ' $a$ ' and bigger triangle be 'b'.
$\Rightarrow \frac{ A }{ B }=\frac{ b ^2}{ a ^2}(\text { By theorem })$
And $a =12 cm, A =225 sq \cdot cm , B=81 sq \cdot cm$......(Given)
$\Rightarrow \frac{225}{81}=\frac{b^2}{12^2}$
$\Rightarrow b^2=\frac{225 \times 144}{81} $
$\Rightarrow b=\sqrt{ } 400 $
$\Rightarrow b=20 cm$
$\Rightarrow \frac{ A }{ B }=\frac{ b ^2}{ a ^2}(\text { By theorem })$
And $a =12 cm, A =225 sq \cdot cm , B=81 sq \cdot cm$......(Given)
$\Rightarrow \frac{225}{81}=\frac{b^2}{12^2}$
$\Rightarrow b^2=\frac{225 \times 144}{81} $
$\Rightarrow b=\sqrt{ } 400 $
$\Rightarrow b=20 cm$
Question 62 Marks
If ΔABC ~ ΔPQR, A (ΔABC) = 80, A(ΔPQR) = 125, then fill in the blanks.
$\frac{ A (\Delta ABC )}{ A (\Delta \ldots)}=\frac{80}{125} \quad \therefore \frac{ AB }{ PQ }=\frac{\square}{\square}$
$\frac{ A (\Delta ABC )}{ A (\Delta \ldots)}=\frac{80}{125} \quad \therefore \frac{ AB }{ PQ }=\frac{\square}{\square}$
Answer
View full question & answer→$\because \triangle ABC \sim \triangle PQR \Rightarrow \frac{ A (\triangle ABC )}{ A (\Delta PQR )}=\frac{80}{125}(\because A (\triangle PQR )=125 \text { is given })$
$\Rightarrow \frac{ A (\triangle ABC )}{ A (\triangle PQR )}=\frac{ AB ^2}{ PQ ^2}$
$\Rightarrow \frac{ AB }{ PQ }=\sqrt{\frac{A(\Delta ABC )}{ A (\triangle PQR )}}$
$\Rightarrow \frac{ AB }{ PQ }=\frac{4}{5}$
$\Rightarrow \frac{ A (\triangle ABC )}{ A (\triangle PQR )}=\frac{ AB ^2}{ PQ ^2}$
$\Rightarrow \frac{ AB }{ PQ }=\sqrt{\frac{A(\Delta ABC )}{ A (\triangle PQR )}}$
$\Rightarrow \frac{ AB }{ PQ }=\frac{4}{5}$
Question 72 Marks
In the figure, in ΔABC, point D on side BC is such that, ∠BAC = ∠ADC.
Prove that, $\mathrm{CA}^2=\mathrm{CB} \times \mathrm{CD}$
Prove that, $\mathrm{CA}^2=\mathrm{CB} \times \mathrm{CD}$
Answer
View full question & answer→In Δ BAC & Δ ADC⇒ ∠ BAC ≅ ∠ ADC ……(Given)
And, ∠ACB ≅ ∠DCA ……(common)
⇒ Δ BAC ~ Δ ADC (By AA Test)
$\Rightarrow \frac{ CA }{ CD }=\frac{ CB }{ CA }$ (corresponding sides are proportional)
⇒ $\mathrm{CA}^2=\mathrm{CB} \times \mathrm{CD}$
And, ∠ACB ≅ ∠DCA ……(common)
⇒ Δ BAC ~ Δ ADC (By AA Test)
$\Rightarrow \frac{ CA }{ CD }=\frac{ CB }{ CA }$ (corresponding sides are proportional)
⇒ $\mathrm{CA}^2=\mathrm{CB} \times \mathrm{CD}$
Question 82 Marks
In the figure, seg AC and seg BD intersect each other in point P and $\frac{AP}{CP} = \frac{BP}{DP}$ Prove that, ΔABP ~ ΔCDP
Answer
In Δ APB & Δ CPD
⇒ $\frac{AP}{CP} = \frac{BP}{DP}$ ……(Given)
And, ∠APB = ∠DPC (vertically opposite angles)
⇒ Δ APB ~ Δ CPD (By SAS Test)
View full question & answer→In Δ APB & Δ CPD
⇒ $\frac{AP}{CP} = \frac{BP}{DP}$ ……(Given)
And, ∠APB = ∠DPC (vertically opposite angles)
⇒ Δ APB ~ Δ CPD (By SAS Test)
Question 92 Marks
□ ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that DE × BE = CE × TE.
Answer
View full question & answer→In Δ CED and ΔBET⇒ ∠ CED≅ ∠ BET (opposite angles)
⇒ ∠ CDE≅ ∠ BTE (Alternate angles)
(∵ AB||DC ⇒ BT||DC, as BT is extension to AB)
⇒ Δ CED ~ Δ BET (By AA Test)
$\Rightarrow \frac{ CE }{ DE }=\frac{ BE }{ TE }$ (corresponding sides are proportional)
⇒ DE× BE = CE×TE
⇒ ∠ CDE≅ ∠ BTE (Alternate angles)
(∵ AB||DC ⇒ BT||DC, as BT is extension to AB)
⇒ Δ CED ~ Δ BET (By AA Test)
$\Rightarrow \frac{ CE }{ DE }=\frac{ BE }{ TE }$ (corresponding sides are proportional)
⇒ DE× BE = CE×TE
Question 102 Marks
In trapezium ABCD, (Figure 1.60) side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15 then find OD.
Answer
View full question & answer→In Δ AOB and ΔCOD⇒ ∠ AOB≅ ∠ COD (opposite angles)
⇒ ∠ CDO≅ ∠ ABO (Alternate angles ∵ AB||DC)
⇒ Δ AOB ~ Δ COD (By AA Test)
$\Rightarrow \frac{ AB }{ DC }=\frac{ OB }{ OD } \text { (corresponding sides are proportional) } $
$ \Rightarrow OD =\frac{ OB \times DC }{ AB } $
$ \Rightarrow O D=\frac{15 \times 6}{20} $
$ \Rightarrow O D=4.5$
⇒ ∠ CDO≅ ∠ ABO (Alternate angles ∵ AB||DC)
⇒ Δ AOB ~ Δ COD (By AA Test)
$\Rightarrow \frac{ AB }{ DC }=\frac{ OB }{ OD } \text { (corresponding sides are proportional) } $
$ \Rightarrow OD =\frac{ OB \times DC }{ AB } $
$ \Rightarrow O D=\frac{15 \times 6}{20} $
$ \Rightarrow O D=4.5$
Question 112 Marks
In ΔABC, AP ⊥ BC, BQ ⊥ AC B- P-C, A-Q - C then prove that, ΔCPA ~ ΔCQB. If AP = 7, BQ = 8, BC = 12 then find AC.
Answer
View full question & answer→From fig.⇒ ∠ APC≅ ∠ BQC (∵ AP⊥ BC and BQ⊥ AC)
⇒ Also, ∠ ACP≅ ∠ BCQ (Common)
⇒ Δ CPA~Δ CQB (By AA Test)
$\Rightarrow \frac{A P}{B Q}=\frac{A C}{B C} $
$ \Rightarrow A C=\frac{A P \times B C}{B Q} $
$ \Rightarrow A C=\frac{7 \times 12}{8}$
⇒AC = 10.5
⇒ Also, ∠ ACP≅ ∠ BCQ (Common)
⇒ Δ CPA~Δ CQB (By AA Test)
$\Rightarrow \frac{A P}{B Q}=\frac{A C}{B C} $
$ \Rightarrow A C=\frac{A P \times B C}{B Q} $
$ \Rightarrow A C=\frac{7 \times 12}{8}$
⇒AC = 10.5
Question 122 Marks
As shown in figure 1.57, two poles of height 8 m and 4 m are perpendicular to the ground. If the length of shadow of smaller pole due to sunlight is 6 m then how long will be the shadow of the bigger pole at the same time?
Answer
View full question & answer→∵ the shadows are measured at the same time⇒ angle of of elevation will be equal for both the pole
⇒ Δ PQR~Δ ABC ………(By AA Test)
$\Rightarrow \frac{ PR }{ AC }=\frac{ QR }{ BC } $
$ \Rightarrow BC =\frac{ QR \times AC }{ PR } $
$ \Rightarrow x =\frac{6 \times 8}{4}$
⇒ x = 12 m
⇒ Δ PQR~Δ ABC ………(By AA Test)
$\Rightarrow \frac{ PR }{ AC }=\frac{ QR }{ BC } $
$ \Rightarrow BC =\frac{ QR \times AC }{ PR } $
$ \Rightarrow x =\frac{6 \times 8}{4}$
⇒ x = 12 m
Question 132 Marks
In Δ ABC, seg BD bisects ∠ ABC. If AB = x, BC = x + 5, AD = x – 2, DC = x + 2, then find the value of x.
Answer
View full question & answer→Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides
$\Rightarrow \frac{A D}{D C}=\frac{A B}{B C}$
$\Rightarrow \frac{x-2}{x+2}=\frac{x}{x+5}$
$\Rightarrow x ( x +2)=( x -2)( x +5)$
$\Rightarrow x ^2+2 x = x ^2-2 x +5 x -10$
$\Rightarrow x ^2+2 x - x ^2+2 x -5 x +10=0$
$\Rightarrow x =10$
$\Rightarrow \frac{A D}{D C}=\frac{A B}{B C}$
$\Rightarrow \frac{x-2}{x+2}=\frac{x}{x+5}$
$\Rightarrow x ( x +2)=( x -2)( x +5)$
$\Rightarrow x ^2+2 x = x ^2-2 x +5 x -10$
$\Rightarrow x ^2+2 x - x ^2+2 x -5 x +10=0$
$\Rightarrow x =10$
Question 142 Marks
In Δ LMN, ray MT bisects ∠ LMN If LM = 6, MN = 10, TN = 8, then find LT.
Answer
View full question & answer→Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.
$\Rightarrow \frac{ LT }{ TN }=\frac{ LM }{ MN }$
$\Rightarrow LT =\frac{ LM \times TN }{ MN } $
$ \Rightarrow LT =\frac{6 \times 8}{10}$
⇒ LT = 4.8
$\Rightarrow \frac{ LT }{ TN }=\frac{ LM }{ MN }$
$\Rightarrow LT =\frac{ LM \times TN }{ MN } $
$ \Rightarrow LT =\frac{6 \times 8}{10}$
⇒ LT = 4.8
Question 152 Marks
In figure 1.41, if AB || CD || FE then find x and AE.
Answer
View full question & answer→Theorem: The ratio of the intercepts made on a transversal by three parallel lines is equal to the ratio of the corresponding intercepts made on any other transversal by the same parallel lines.
$\Rightarrow \frac{ BD }{ DF }=\frac{ AC }{ CE } $
$ \Rightarrow \frac{8}{4}=\frac{12}{ x }$
$ \Rightarrow x =\frac{12 \times 4}{8}$
⇒ x = 6
Now,AE = AC + CE
= 12 + x
= 12 + 6
⇒ AE = 18
$\Rightarrow \frac{ BD }{ DF }=\frac{ AC }{ CE } $
$ \Rightarrow \frac{8}{4}=\frac{12}{ x }$
$ \Rightarrow x =\frac{12 \times 4}{8}$
⇒ x = 6
Now,AE = AC + CE
= 12 + x
= 12 + 6
⇒ AE = 18
Question 162 Marks
Find QP using given information in the figure.
Answer
View full question & answer→Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.And ∵ NQ is angle bisector of ∠N
$\Rightarrow \frac{M Q}{Q P}=\frac{M N}{N P} $
$ \Rightarrow \frac{14}{Q P}=\frac{25}{40} $
$\Rightarrow Q P=\frac{14 \times 40}{25}$
⇒ QP = 22.4
$\Rightarrow \frac{M Q}{Q P}=\frac{M N}{N P} $
$ \Rightarrow \frac{14}{Q P}=\frac{25}{40} $
$\Rightarrow Q P=\frac{14 \times 40}{25}$
⇒ QP = 22.4
Question 172 Marks
In trapezium ABCD, side AB || side PQ || side DC, AP = 15, PD = 12, QC = 14, find BQ.
Answer
View full question & answer→By Basic Proportionality Theorem(Theorem : If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.)
$\Rightarrow \frac{ AP }{ PD }=\frac{ BQ }{ QC }$
$\Rightarrow \frac{15}{12}=\frac{ BQ }{14} $
$\Rightarrow BQ =\frac{15 \times 14}{12}$
⇒ BQ = 17.5
$\Rightarrow \frac{ AP }{ PD }=\frac{ BQ }{ QC }$
$\Rightarrow \frac{15}{12}=\frac{ BQ }{14} $
$\Rightarrow BQ =\frac{15 \times 14}{12}$
⇒ BQ = 17.5
Question 182 Marks
Measures of some angles in the figure are given. Prove that $\frac{AP}{PB} = \frac{AQ}{QC}$
Answer
View full question & answer→Here, PQ||BC (∵ ∠ APQ≅ ∠ ABC)(PROPERTY: If a transversal intersects two lines so that corresponding angles are congruent, then the lines are parallel)
∴ By Basic Proportionality Theorem
(Theorem : If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.)
$\Rightarrow \frac{AP}{PB} = \frac{AQ}{QC}$
∴ By Basic Proportionality Theorem
(Theorem : If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.)
$\Rightarrow \frac{AP}{PB} = \frac{AQ}{QC}$
Question 192 Marks
In Δ MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7 MQ = 2.5 then find QP.
Answer
View full question & answer→Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.
$\Rightarrow \frac{ MQ }{ QP }=\frac{ MN }{ NP }$
$ \Rightarrow \frac{2.5}{ QP }=\frac{5}{7} $
$ \Rightarrow QP \times 5=2.5 \times 7 $
$ \Rightarrow QP =\frac{2.5 \times 7}{5}$
⇒ QP = 3.5
$\Rightarrow \frac{ MQ }{ QP }=\frac{ MN }{ NP }$
$ \Rightarrow \frac{2.5}{ QP }=\frac{5}{7} $
$ \Rightarrow QP \times 5=2.5 \times 7 $
$ \Rightarrow QP =\frac{2.5 \times 7}{5}$
⇒ QP = 3.5
Question 202 Marks
In figure 1.81, the vertices of square DEFG are on the sides of ΔABC, ∠A = 90°. Then prove that $D E^2=B D \times E C$
(Hint : Show that ΔGBD is similar to ΔDFE. Use GD = FE = DE.)
(Hint : Show that ΔGBD is similar to ΔDFE. Use GD = FE = DE.)
Answer
View full question & answer→Proof:In □ DEFG is a square
⇒ GF||DE
⇒ GF||BC
Now,In Δ AGF and Δ DBG
⇒ ∠ AGF≅ ∠ DBG (corresponding angles)
⇒ ∠ GDB≅ ∠ FAG (Both are 90° )
⇒ Δ AGF~Δ DBG ……(1) (AA similarity)
Now,In Δ AGF and Δ EFC
⇒ ∠ AFG≅ ∠ ECF (corresponding angles)
⇒ ∠ GAF≅ ∠ FEC (Both are 90° )
⇒ Δ AGF~Δ EFC ……(2) (AA similarity)
From (1) & (2), we have-
⇒ Δ EFC~Δ DBG
$\Rightarrow \frac{ EF }{ BD }=\frac{ EC }{ DG }$
⇒ EF× DG = BD× EC
Now,∵ DEFG is a square
⇒ DE = EF = DG
⇒ DE× DE = BD× EC
⇒ $DE^2$= BD× EC
⇒ GF||DE
⇒ GF||BC
Now,In Δ AGF and Δ DBG
⇒ ∠ AGF≅ ∠ DBG (corresponding angles)
⇒ ∠ GDB≅ ∠ FAG (Both are 90° )
⇒ Δ AGF~Δ DBG ……(1) (AA similarity)
Now,In Δ AGF and Δ EFC
⇒ ∠ AFG≅ ∠ ECF (corresponding angles)
⇒ ∠ GAF≅ ∠ FEC (Both are 90° )
⇒ Δ AGF~Δ EFC ……(2) (AA similarity)
From (1) & (2), we have-
⇒ Δ EFC~Δ DBG
$\Rightarrow \frac{ EF }{ BD }=\frac{ EC }{ DG }$
⇒ EF× DG = BD× EC
Now,∵ DEFG is a square
⇒ DE = EF = DG
⇒ DE× DE = BD× EC
⇒ $DE^2$= BD× EC
Question 212 Marks
In □ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that $\frac{A P}{P D}=\frac{P C}{B P}$
Answer
View full question & answer→In Δ APD and ΔCPB⇒ ∠ APD≅ ∠ CPB (opposite angles)
⇒ ∠ ADP≅ ∠ PBC (Alternate angles ∵ AD||BC)
⇒ Δ APD ~ Δ CPB (By AA Test)
$\Rightarrow \frac{ AP }{ PC }=\frac{ PD }{ BP } \text { (corresponding sides are proportional) }$
$ \Rightarrow \frac{ AP }{ PD }=\frac{ PC }{ BP }$
⇒ ∠ ADP≅ ∠ PBC (Alternate angles ∵ AD||BC)
⇒ Δ APD ~ Δ CPB (By AA Test)
$\Rightarrow \frac{ AP }{ PC }=\frac{ PD }{ BP } \text { (corresponding sides are proportional) }$
$ \Rightarrow \frac{ AP }{ PD }=\frac{ PC }{ BP }$
Question 222 Marks
In ΔPQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR.
Complete the proof by filling in the boxes. In $\triangle PMQ$, ray MX is bisector of $\angle PMQ$.
$\therefore\frac{\square}{\square}=\frac{\square}{\square}......... (I)$ theorem of angle bisector.
In $\triangle PMR$, ray MY is bisector of $\angle PMR$.
$\therefore\frac{\square}{\square}=\frac{\square}{\square}.........(II)$ theorem of angle bisector.
But $\frac{M P}{M Q}=\frac{M P}{M R}$ $\qquad$
$\therefore \frac{ PX }{ XQ }=\frac{ PY }{ YR }$ $\qquad$
$\therefore XY||QR......$ converse of basic proportionality theorem.
Complete the proof by filling in the boxes. In $\triangle PMQ$, ray MX is bisector of $\angle PMQ$.
$\therefore\frac{\square}{\square}=\frac{\square}{\square}......... (I)$ theorem of angle bisector.
In $\triangle PMR$, ray MY is bisector of $\angle PMR$.
$\therefore\frac{\square}{\square}=\frac{\square}{\square}.........(II)$ theorem of angle bisector.
But $\frac{M P}{M Q}=\frac{M P}{M R}$ $\qquad$
$\therefore \frac{ PX }{ XQ }=\frac{ PY }{ YR }$ $\qquad$
$\therefore XY||QR......$ converse of basic proportionality theorem.
Answer
View full question & answer→$\therefore \frac{ PM }{ MQ }=\frac{ PX }{ XQ }............ (I)$ theorem of angle bisector.
AND
$\therefore \frac{ PM }{ MR }=\frac{ PY }{ YR }.........(II)$ theorem of angle bisector.
AND
$\therefore \frac{ PM }{ MR }=\frac{ PY }{ YR }.........(II)$ theorem of angle bisector.
Question 232 Marks
If figure 1.13 BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find $\frac{ A (\Delta ABC )}{ A (\Delta ADB )}$
Answer
View full question & answer→Here, $\triangle ABC$ and $\triangle ADB$ has common Base. $\therefore \frac{\operatorname{Ar}(\triangle ABC )}{\operatorname{Ar}(\triangle ADB )}=\frac{\text { height of } \triangle ABC }{\text { height of } \triangle ADB }$
(PROPERTY: Areas of triangles with equal bases are proportional to their corresponding heights.)
$\Rightarrow \frac{\operatorname{Ar}(\triangle ABC )}{\operatorname{Ar}(\triangle ADB )}=\frac{ BC }{ AD } $
$=\frac{4}{8}$
$=\frac{1}{2}$
(PROPERTY: Areas of triangles with equal bases are proportional to their corresponding heights.)
$\Rightarrow \frac{\operatorname{Ar}(\triangle ABC )}{\operatorname{Ar}(\triangle ADB )}=\frac{ BC }{ AD } $
$=\frac{4}{8}$
$=\frac{1}{2}$
Question 242 Marks
Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.
Answer
View full question & answer→We know that area of triangle $=\frac{1}{2} \times$ Base $\times$ Height
$\Rightarrow \text { Area }(\text { triangle } 1)=\frac{1}{2} \times 9 \times 5 $
$=\frac{45}{2} $
$\Rightarrow \text { Area (triangle 2) }=\frac{1}{2} \times 10 \times 6$
$=30$
$\therefore$ the ratio of areas of these triangles will be $=\frac{\text { Area }(\text { triangle } 1)}{\text { Area(triangle } 2)}$
$=\frac{\frac{45}{2}}{30} $
$=\frac{45}{2} \times \frac{1}{30}$
$=\frac{3}{4}$
$\Rightarrow \text { Area }(\text { triangle } 1)=\frac{1}{2} \times 9 \times 5 $
$=\frac{45}{2} $
$\Rightarrow \text { Area (triangle 2) }=\frac{1}{2} \times 10 \times 6$
$=30$
$\therefore$ the ratio of areas of these triangles will be $=\frac{\text { Area }(\text { triangle } 1)}{\text { Area(triangle } 2)}$
$=\frac{\frac{45}{2}}{30} $
$=\frac{45}{2} \times \frac{1}{30}$
$=\frac{3}{4}$
Question 252 Marks
In figure 1.75, A – D – C and B – E – C seg DE || side AB If AD = 5, DC = 3, BC = 6.4 then find BE.
Answer
View full question & answer→By Basic Proportionality Theorem-
$\Rightarrow \frac{C D}{D A}=\frac{C E}{E B}$
$ \Rightarrow \frac{3}{5}=\frac{6.4-x}{x}$
⇒ 3x = 32-5x
⇒ 8x = 32
⇒ x = 4 = BE
$\Rightarrow \frac{C D}{D A}=\frac{C E}{E B}$
$ \Rightarrow \frac{3}{5}=\frac{6.4-x}{x}$
⇒ 3x = 32-5x
⇒ 8x = 32
⇒ x = 4 = BE
Question 262 Marks
ΔMNT ~ ΔQRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio $\frac{ A (\Delta MNT )}{ A (\Delta QRS )}$
Answer
View full question & answer→$\frac{ A (\Delta MNT )}{ A (\Delta QRS )}=\frac{(\text { altitude from } T )^2}{(\text { altitude frm } S)^2}$
$ =\frac{5^2}{9^2} $
$ =\frac{25}{81}$
$ =\frac{5^2}{9^2} $
$ =\frac{25}{81}$
Question 272 Marks
In trapezium $A B C D$, side $A B \|$ side $C D$, diagonal $A C$ and $B D$ intersect each other at point $P$. Then prove that $\frac{A(\Delta \mathrm{ABP})}{\mathrm{A}(\Delta \mathrm{CPD})}=\frac{\mathrm{AB}^2}{\mathrm{CD}^2}$.
Answer
View full question & answer→In trapezium $\mathrm{ABCD}$ side $\mathrm{AB} \|$ side $\mathrm{CD}$ In $\triangle \mathrm{APB}$ and $\triangle \mathrm{CPD}$
$\angle \mathrm{PAB} \cong \angle \mathrm{PCD}$ alternate angles
$\angle \mathrm{APB} \cong \angle \mathrm{CPD}$ opposite angles
$\therefore \triangle \mathrm{APB} \sim \triangle \mathrm{CPD}$ AA test of similarity
$\frac{\mathrm{A}(\Delta \mathrm{APB})}{\mathrm{A}(\Delta \mathrm{CPD})}=\frac{\mathrm{AB}^2}{\mathrm{CD}^2} \ldots \ldots$ theorem of areas of similar triangles
$\angle \mathrm{PAB} \cong \angle \mathrm{PCD}$ alternate angles
$\angle \mathrm{APB} \cong \angle \mathrm{CPD}$ opposite angles
$\therefore \triangle \mathrm{APB} \sim \triangle \mathrm{CPD}$ AA test of similarity
$\frac{\mathrm{A}(\Delta \mathrm{APB})}{\mathrm{A}(\Delta \mathrm{CPD})}=\frac{\mathrm{AB}^2}{\mathrm{CD}^2} \ldots \ldots$ theorem of areas of similar triangles
Question 282 Marks
Ratio of corresponging sides of two similar triangles is $2:5,$ If the area of the small triangle is $64$ sq.cm. then what is the area of the bigger triangle ?
Answer
View full question & answer→Assume that $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$.
$\triangle \mathrm{ABC}$ is smaller and $\triangle \mathrm{PQR}$ is bigger triangle.
$ \therefore \frac{\mathrm{A}(\triangle \mathrm{ABC})}{\mathrm{A}(\triangle \mathrm{PQR})}=\frac{(2)^2}{(5)^2}=\frac{4}{25} \ldots \ldots . . . \text { ratio of areas of similar triangles }$
$\therefore \frac{64}{\mathrm{~A}(\Delta \mathrm{PQR})}=\frac{4}{25}$
$4 \times \mathrm{A}(\triangle \mathrm{PQR})=64 \times 25$
$\mathrm{~A}(\Delta \mathrm{PQR})=\frac{64 \times 25}{4}=400$
$\therefore \text { area of bigger triangle }=400 \mathrm{sq} . \mathrm{cm} . $
$\triangle \mathrm{ABC}$ is smaller and $\triangle \mathrm{PQR}$ is bigger triangle.
$ \therefore \frac{\mathrm{A}(\triangle \mathrm{ABC})}{\mathrm{A}(\triangle \mathrm{PQR})}=\frac{(2)^2}{(5)^2}=\frac{4}{25} \ldots \ldots . . . \text { ratio of areas of similar triangles }$
$\therefore \frac{64}{\mathrm{~A}(\Delta \mathrm{PQR})}=\frac{4}{25}$
$4 \times \mathrm{A}(\triangle \mathrm{PQR})=64 \times 25$
$\mathrm{~A}(\Delta \mathrm{PQR})=\frac{64 \times 25}{4}=400$
$\therefore \text { area of bigger triangle }=400 \mathrm{sq} . \mathrm{cm} . $
Question 292 Marks
$\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}, \mathrm{A}(\triangle \mathrm{ABC})=16, \mathrm{~A}(\triangle \mathrm{PQP})=25$, then find the value of ratio $\frac{\mathrm{AB}}{\mathrm{PQ}}$.
Answer
View full question & answer→$: \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$
$\therefore \frac{\mathrm{A}(\triangle \mathrm{ABC})}{\mathrm{A}(\Delta \mathrm{PQR})}=\frac{\mathrm{AB}^2}{\mathrm{PQ}^2}\text{ .......... theorem of areas of similar triangles}$
$\therefore \frac{16}{25}=\frac{\mathrm{AB}^2}{\mathrm{PQ}^2} \quad \therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{4}{5}\text{.......... taking square roots} $
theorem of areas of similar triangles taking square roots
$\therefore \frac{\mathrm{A}(\triangle \mathrm{ABC})}{\mathrm{A}(\Delta \mathrm{PQR})}=\frac{\mathrm{AB}^2}{\mathrm{PQ}^2}\text{ .......... theorem of areas of similar triangles}$
$\therefore \frac{16}{25}=\frac{\mathrm{AB}^2}{\mathrm{PQ}^2} \quad \therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{4}{5}\text{.......... taking square roots} $
theorem of areas of similar triangles taking square roots
Question 302 Marks
Can we say that the two triangles in figure $1.52$ similar, according to information given? If yes, by which test ?
Answer
View full question & answer→$\Delta \mathrm{XYZ}$ and $\Delta \mathrm{MNP}$,
$\frac{X Y}{M N}=\frac{14}{21}=\frac{2}{3},$
$\frac{Y Z}{N P}=\frac{20}{30}=\frac{2}{3}$
It is given that $\angle \mathrm{Z} \cong \angle \mathrm{P}$.
But $\angle \mathrm{Z}$ and $\angle \mathrm{P}$ are not included angles by sides which are in proportion.
: $\Delta \mathrm{XYZ}$ and $\Delta \mathrm{MNP}$ can not be said to be similar.
$\frac{X Y}{M N}=\frac{14}{21}=\frac{2}{3},$
$\frac{Y Z}{N P}=\frac{20}{30}=\frac{2}{3}$
It is given that $\angle \mathrm{Z} \cong \angle \mathrm{P}$.
But $\angle \mathrm{Z}$ and $\angle \mathrm{P}$ are not included angles by sides which are in proportion.
: $\Delta \mathrm{XYZ}$ and $\Delta \mathrm{MNP}$ can not be said to be similar.
Question 312 Marks
Given : line $l \|$ line $m \|$ line $n$ $t_1$ and $t_2$ are transversals. Transversal $t_1$ intersects the lines in points A, B, C and $\mathrm{t}_2$ intersects the lines in points $\mathrm{P}$, Q, R. To prove : $\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{PQ}}{\mathrm{QR}}$
View full question & answer→Question 322 Marks
If $\triangle ABC \sim \triangle DEF$ such that the area of $\triangle ABC$ is $9 cm^2$ and the area of $\triangle D E F$ is $16 cm^2$. If $B C=2.1$ cm . Find length of EF.
Answer
View full question & answer→2.8
Question 332 Marks
$\triangle D E F \sim \triangle M N K$, If $D E=5$ and $M N=6$, then find the value of $A (\triangle DEF ): A (\triangle MNK )$
Answer
View full question & answer→$25: 36$
Question 342 Marks
Let $X$ be any point on side $BC$ of $\Delta{ABC}$. seg $XM ||$ side $AB$ and seg $XN ||$ side $CA. M-N-T, T-B-X.$ Prove that : $TX^2 = TB.TC.$
Answer
View full question & answer→self
Question 352 Marks
A vertical pole of a length 6 m casts a shadow of 4 m long on the ground. At the same time a tower casts a shadow 28 m long. Find the height of the tower.
Answer
View full question & answer→42 m
Question 362 Marks
In $\Delta$XYZ, XY = YZ. Ray YM bisects $\angle$XYZ. X-M-Z. Prove that M is midpoint of seg XZ.
Answer
View full question & answer→self
Question 372 Marks
In $\Delta$PQR, seg RS is bisector of $\angle P R Q$ PS = 6, SQ = 8, PR = 15. Find QR.
Answer
View full question & answer→ 20 units
Question 382 Marks
In $\Delta$DEF, line PQ || side EF. DQ = 1.8, QF = 5.4, PE = 7.2. Find DE.
Answer
View full question & answer→9.6 units
Question 392 Marks
The ratio of the areas of two triangles with equal height is $3: 2$. The base of the larger triangle is 18 cm . Find the corresponding base of the smaller triangle.
Answer
View full question & answer→12 cm
Question 402 Marks
Prove The Theorem : If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Answer
In figure, line l interesects the side AB and side AC of $\triangle ABC$ in the points P and Q respectively and $\frac{A P}{P B}=\frac{A Q}{Q C}$, hence line $l \|$seg BC.
This theorem can be proved by indirect method.
View full question & answer→In figure, line l interesects the side AB and side AC of $\triangle ABC$ in the points P and Q respectively and $\frac{A P}{P B}=\frac{A Q}{Q C}$, hence line $l \|$seg BC.
This theorem can be proved by indirect method.
Question 412 Marks
Prove The Theorem : The ratio of the intercepts made on a transversal by three parallel lines is equal to the ratio of the corrosponding intercepts made on any other transversal by the same parallel lines.
Answer
View full question & answer→Given : line $l \|$ line $m \|$ line $n$ $t_1$ and $t_2$ are transversals. Transversal $t_1$ intersects the lines in points A, B, C and $\mathrm{t}_2$ intersects the lines in points $\mathrm{P}$, Q, R. To prove : $\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{PQ}}{\mathrm{QR}}$
Question 422 Marks
View full question & answer→
Question 432 Marks
ΔMNT ~ ΔQRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio $\frac{ A (\Delta MNT )}{ A (\Delta QRS )}$
Answer
View full question & answer→$\begin{array}{l}\frac{ A (\Delta MNT )}{ A (\Delta QRS )}=\frac{(\text { altitude from } T )^2}{\text { (altitude frm } S ^2} \\ =\frac{5^2}{9^2} \\ =\frac{25}{81}\end{array}$
Question 442 Marks
Areas of two similar triangles are 225 sq.cm & 81 sq.cm. If a side of the smaller triangle is 12 cm, then find corresponding side of the bigger triangle.
Question 452 Marks
In the figure, in ΔABC, point D on side BC is such that, ∠BAC = ∠ADC.
Prove that, CA2 = CB × CD
Prove that, CA2 = CB × CD
Answer
$\begin{array}{l}\text { In } \triangle BAC \& \triangle ADC \\
\Rightarrow \angle BAC \cong \angle ADC \ldots \ldots \text { (Given) } \\
\text { And, } \angle ACB \cong \angle DCA \ldots \ldots \text { (common) } \\
\Rightarrow \triangle BAC \sim \triangle ADC \text { (By AA Test) } \\
\Rightarrow \frac{ CA }{ CD }=\frac{ CB }{ CA } \text { (corresponding sides are proportional) } \\
\Rightarrow CA ^2= CB \times CD
\end{array}$
(common)
View full question & answer→$\begin{array}{l}\text { In } \triangle BAC \& \triangle ADC \\
\Rightarrow \angle BAC \cong \angle ADC \ldots \ldots \text { (Given) } \\
\text { And, } \angle ACB \cong \angle DCA \ldots \ldots \text { (common) } \\
\Rightarrow \triangle BAC \sim \triangle ADC \text { (By AA Test) } \\
\Rightarrow \frac{ CA }{ CD }=\frac{ CB }{ CA } \text { (corresponding sides are proportional) } \\
\Rightarrow CA ^2= CB \times CD
\end{array}$
(common)
Question 462 Marks
Answer
View full question & answer→In $\triangle APB \& \triangle CPD$
$\Rightarrow \frac{ AP }{ CP }=\frac{ BP }{ DP } \ldots \ldots$ (Given)
And, $\angle APB =\angle DPC$ (vertically opposite angles)
$\Rightarrow \triangle APB \sim \Delta CPD ($ By SAS Test)
$\Rightarrow \frac{ AP }{ CP }=\frac{ BP }{ DP } \ldots \ldots$ (Given)
And, $\angle APB =\angle DPC$ (vertically opposite angles)
$\Rightarrow \triangle APB \sim \Delta CPD ($ By SAS Test)
Question 472 Marks
$J A B C D$ is a parallelogram. Point $E$ is on side $B C$. Line $D E$ intersects ray $A B$ in point $T$. Prove that $D E \times B E=C E \times T E$.
Answer
$\begin{array}{l}\text { In } \triangle CED \text { and } \triangle BET \\ \Rightarrow \angle CED \cong \angle BET \text { (opposite angles) } \\ \Rightarrow \angle CDE \cong \angle BTE \text { (Alternate angles) } \\ (\because AB \| DC \Rightarrow BT \| DC \text {, as } BT \text { is extension to } AB ) \\ \Rightarrow \triangle CED \sim \triangle BET \text { (By AA Test) } \\ \Rightarrow \frac{ CE }{ DE }=\frac{ BE }{ TE } \text { (corresponding sides are proportional) } \\ \Rightarrow DE \times BE = CE \times TE \end{array}$
View full question & answer→$\begin{array}{l}\text { In } \triangle CED \text { and } \triangle BET \\ \Rightarrow \angle CED \cong \angle BET \text { (opposite angles) } \\ \Rightarrow \angle CDE \cong \angle BTE \text { (Alternate angles) } \\ (\because AB \| DC \Rightarrow BT \| DC \text {, as } BT \text { is extension to } AB ) \\ \Rightarrow \triangle CED \sim \triangle BET \text { (By AA Test) } \\ \Rightarrow \frac{ CE }{ DE }=\frac{ BE }{ TE } \text { (corresponding sides are proportional) } \\ \Rightarrow DE \times BE = CE \times TE \end{array}$
Question 482 Marks
Id trapezium ABCD (adjoining figure), side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15 then find OD.
Answer
$\begin{array}{l}\text { In } \triangle AOB \text { and } \triangle COD \\ \Rightarrow \angle AOB \angle COD \text { (opposite angles) } \\ \Rightarrow \angle CDO \cong \angle ABO \text { (Alternate angles } \because AB \| DC ) \\ \Rightarrow \triangle AOB \sim \triangle COD \text { (By AA Test) } \\ \Rightarrow \frac{ AB }{ DC }=\frac{ OB }{ OD } \text { (corresponding sides are proportional) } \\ \Rightarrow OD =\frac{ OB \times DC }{ AB } \\ \Rightarrow OD =\frac{15 \times 6}{20} \\ \Rightarrow OD =4.5\end{array}$
View full question & answer→$\begin{array}{l}\text { In } \triangle AOB \text { and } \triangle COD \\ \Rightarrow \angle AOB \angle COD \text { (opposite angles) } \\ \Rightarrow \angle CDO \cong \angle ABO \text { (Alternate angles } \because AB \| DC ) \\ \Rightarrow \triangle AOB \sim \triangle COD \text { (By AA Test) } \\ \Rightarrow \frac{ AB }{ DC }=\frac{ OB }{ OD } \text { (corresponding sides are proportional) } \\ \Rightarrow OD =\frac{ OB \times DC }{ AB } \\ \Rightarrow OD =\frac{15 \times 6}{20} \\ \Rightarrow OD =4.5\end{array}$
Question 492 Marks
Answer
View full question & answer→From fig.
$\begin{array}{l}
\Rightarrow \angle APC \cong \angle BQC (\because AP \perp BC \text { and } BQ \perp AC ) \\
\Rightarrow Also , \angle ACP \cong \angle BCQ \text { (Common) } \\
\Rightarrow \triangle CPA \sim \triangle CQB \text { (By AA Test) } \\
\Rightarrow \frac{ AP }{ BQ }=\frac{ AC }{ BC } \\
\Rightarrow AC =\frac{ AP \times BC }{ BQ } \\
\Rightarrow AC =\frac{7 \times 12}{8} \\
\Rightarrow AC =10.5
\end{array}$
$\begin{array}{l}
\Rightarrow \angle APC \cong \angle BQC (\because AP \perp BC \text { and } BQ \perp AC ) \\
\Rightarrow Also , \angle ACP \cong \angle BCQ \text { (Common) } \\
\Rightarrow \triangle CPA \sim \triangle CQB \text { (By AA Test) } \\
\Rightarrow \frac{ AP }{ BQ }=\frac{ AC }{ BC } \\
\Rightarrow AC =\frac{ AP \times BC }{ BQ } \\
\Rightarrow AC =\frac{7 \times 12}{8} \\
\Rightarrow AC =10.5
\end{array}$
Question 502 Marks
As shown in the adjoining figure, two poles of height 8 m and 4 m are perpendicular to the ground. If the length of shadow of smaller pole due to sunlight is 6 m, then how long will be the shadow of the bigger pole at the same time?
Answer
$\because$ the shadows are measured at the same time
$\Rightarrow$ angle of of elevation will be equal for both the pole
$\begin{array}{l}
\Rightarrow \triangle PQR \sim \triangle ABC \ldots \ldots . . \text { (By AA Test) } \\
\Rightarrow \frac{ PR }{ AC }=\frac{ QR }{ BC } \\
\Rightarrow BC =\frac{ QR \times AC }{ PR } \\
\Rightarrow x =\frac{6 \times q }{4} \\
\Rightarrow x =12 m
\end{array}$
(By AA Test)
View full question & answer→$\because$ the shadows are measured at the same time
$\Rightarrow$ angle of of elevation will be equal for both the pole
$\begin{array}{l}
\Rightarrow \triangle PQR \sim \triangle ABC \ldots \ldots . . \text { (By AA Test) } \\
\Rightarrow \frac{ PR }{ AC }=\frac{ QR }{ BC } \\
\Rightarrow BC =\frac{ QR \times AC }{ PR } \\
\Rightarrow x =\frac{6 \times q }{4} \\
\Rightarrow x =12 m
\end{array}$
(By AA Test)
Question 512 Marks
Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm then what is the corresponding base of the bigger triangle?
Question 522 Marks
Answer
Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.
$\begin{array}{l}
\Rightarrow \frac{A D}{D C}=\frac{A B}{B C} \\
\Rightarrow \frac{x-2}{x+2}=\frac{x}{x+5} \\
\Rightarrow x(x+2)=(x-2)(x+5) \\
\Rightarrow x^2+2 x=x^2-2 x+5 x-10 \\
\Rightarrow x^2+2 x-x^2+2 x-5 x+10=0 \\
\Rightarrow x=10
\end{array}$
View full question & answer→Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.
$\begin{array}{l}
\Rightarrow \frac{A D}{D C}=\frac{A B}{B C} \\
\Rightarrow \frac{x-2}{x+2}=\frac{x}{x+5} \\
\Rightarrow x(x+2)=(x-2)(x+5) \\
\Rightarrow x^2+2 x=x^2-2 x+5 x-10 \\
\Rightarrow x^2+2 x-x^2+2 x-5 x+10=0 \\
\Rightarrow x=10
\end{array}$
Question 532 Marks
Answer
Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.
$\begin{array}{l}
\Rightarrow \frac{ LT }{ TN }=\frac{ LM }{ MN } \\
\Rightarrow LT =\frac{ LM \times TN }{ MN } \\
\Rightarrow LT =\frac{6 \times 8}{10} \\
\Rightarrow LT =4.8
\end{array}$
View full question & answer→Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.
$\begin{array}{l}
\Rightarrow \frac{ LT }{ TN }=\frac{ LM }{ MN } \\
\Rightarrow LT =\frac{ LM \times TN }{ MN } \\
\Rightarrow LT =\frac{6 \times 8}{10} \\
\Rightarrow LT =4.8
\end{array}$
Question 542 Marks
In the adjoining figure, if AB || CD || FE then find x and AE.
Answer
Theorem: The ratio of the intercepts made on a transversal by three parallel lines is equal to the ratio of the corresponding intercepts made on any other transversal by the same parallel lines.
$\begin{array}{l}
\Rightarrow \frac{ BD }{ DF }=\frac{ AC }{ CE } \\
\Rightarrow \frac{8}{4}=\frac{12}{ x } \\
\Rightarrow x =\frac{12 \times 4}{8} \\
\Rightarrow x =6
\end{array}$
Now, $A E=A C+C E$
$\begin{array}{l}
=12+x \\
=12+6 \\
\Rightarrow A E=18
\end{array}$
View full question & answer→Theorem: The ratio of the intercepts made on a transversal by three parallel lines is equal to the ratio of the corresponding intercepts made on any other transversal by the same parallel lines.
$\begin{array}{l}
\Rightarrow \frac{ BD }{ DF }=\frac{ AC }{ CE } \\
\Rightarrow \frac{8}{4}=\frac{12}{ x } \\
\Rightarrow x =\frac{12 \times 4}{8} \\
\Rightarrow x =6
\end{array}$
Now, $A E=A C+C E$
$\begin{array}{l}
=12+x \\
=12+6 \\
\Rightarrow A E=18
\end{array}$
Question 552 Marks
Answer
View full question & answer→Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.
And $\because NQ$ is angle bisector of $\angle N$
$\begin{array}{l}
\Rightarrow \frac{ MQ }{ QP }=\frac{ MN }{ NP } \\
\Rightarrow \frac{14}{ QP }=\frac{25}{40} \\
\Rightarrow Q P=\frac{14 \times 4 0}{25} \\
\Rightarrow Q P=22.4
\end{array}$
And $\because NQ$ is angle bisector of $\angle N$
$\begin{array}{l}
\Rightarrow \frac{ MQ }{ QP }=\frac{ MN }{ NP } \\
\Rightarrow \frac{14}{ QP }=\frac{25}{40} \\
\Rightarrow Q P=\frac{14 \times 4 0}{25} \\
\Rightarrow Q P=22.4
\end{array}$
Question 562 Marks
Answer
View full question & answer→By_Basic Proportionality Theorem
(Theorem : If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.)
$\begin{aligned}
& \frac{ AP }{ PD }=\frac{ BQ }{ QC } \\
\Rightarrow & \frac{15}{12}=\frac{ BQ }{14} \\
\Rightarrow & BQ =\frac{15 \times 14}{12} \\
\Rightarrow & BQ =17.5
\end{aligned}$
(Theorem : If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.)
$\begin{aligned}
& \frac{ AP }{ PD }=\frac{ BQ }{ QC } \\
\Rightarrow & \frac{15}{12}=\frac{ BQ }{14} \\
\Rightarrow & BQ =\frac{15 \times 14}{12} \\
\Rightarrow & BQ =17.5
\end{aligned}$
Question 572 Marks
Answer
View full question & answer→Here, $PQ \| BC (\because \angle APQ \cong \angle ABC )$
(PROPERTY: If a transversal intersects two lines so that corresponding angles are congruent, then the lines are parallel)
$\therefore$ By Basic Proportionality Theorem
(Theorem : If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.)
$\Rightarrow \frac{ AP }{ PB }=\frac{ AQ }{ QC }$
(PROPERTY: If a transversal intersects two lines so that corresponding angles are congruent, then the lines are parallel)
$\therefore$ By Basic Proportionality Theorem
(Theorem : If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.)
$\Rightarrow \frac{ AP }{ PB }=\frac{ AQ }{ QC }$
Question 582 Marks
Answer
View full question & answer→Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.
$\begin{array}{l}
\Rightarrow \frac{ MQ }{ QP }=\frac{ MN }{ NP } \\
\Rightarrow \frac{2.5}{ QP }=\frac{5}{7} \\
\Rightarrow Q P \times 5=2.5 \times 7 \\
\Rightarrow Q P=\frac{2.5 \times 7}{5} \\
\Rightarrow Q P=3.5
\end{array}$
$\begin{array}{l}
\Rightarrow \frac{ MQ }{ QP }=\frac{ MN }{ NP } \\
\Rightarrow \frac{2.5}{ QP }=\frac{5}{7} \\
\Rightarrow Q P \times 5=2.5 \times 7 \\
\Rightarrow Q P=\frac{2.5 \times 7}{5} \\
\Rightarrow Q P=3.5
\end{array}$
Question 592 Marks
Answer
View full question & answer→Proof:In □ DEFG is a square
⇒ GF||DE
⇒ GF||BC
Now,In Δ AGF and Δ DBG
⇒ ∠ AGF≅ ∠ DBG (corresponding angles)
⇒ ∠ GDB≅ ∠ FAG (Both are 90° )
⇒ Δ AGF~Δ DBG ……(1) (AA similarity)
Now,In Δ AGF and Δ EFC
⇒ ∠ AFG≅ ∠ ECF (corresponding angles)
⇒ ∠ GAF≅ ∠ FEC (Both are 90° )
⇒ Δ AGF~Δ EFC ……(2) (AA similarity)
From (1) & (2), we have-
⇒ Δ EFC~Δ DBG
⇒ $\Rightarrow \frac{ EF }{ BD }=\frac{ EC }{ DG }$
⇒ EF× DG = BD× EC
Now,∵ DEFG is a square
⇒ DE = EF = DG
⇒ DE× DE = BD× EC
⇒ DE2 = BD× EC
⇒ GF||DE
⇒ GF||BC
Now,In Δ AGF and Δ DBG
⇒ ∠ AGF≅ ∠ DBG (corresponding angles)
⇒ ∠ GDB≅ ∠ FAG (Both are 90° )
⇒ Δ AGF~Δ DBG ……(1) (AA similarity)
Now,In Δ AGF and Δ EFC
⇒ ∠ AFG≅ ∠ ECF (corresponding angles)
⇒ ∠ GAF≅ ∠ FEC (Both are 90° )
⇒ Δ AGF~Δ EFC ……(2) (AA similarity)
From (1) & (2), we have-
⇒ Δ EFC~Δ DBG
⇒ $\Rightarrow \frac{ EF }{ BD }=\frac{ EC }{ DG }$
⇒ EF× DG = BD× EC
Now,∵ DEFG is a square
⇒ DE = EF = DG
⇒ DE× DE = BD× EC
⇒ DE2 = BD× EC
Question 602 Marks
Answer
Here, $\triangle ABC$ and $\triangle ADB$ has common Base.
$\therefore \frac{\operatorname{Ar}(\triangle ABC )}{\operatorname{Ar}(\triangle ADB )}=\frac{\text { height of } \triangle ABO }{\text { height of } \triangle ADB }$
(PROPERTY: Areas of triangles with equal bases are proportional to their corresponding heights.)
$\begin{aligned}
& \frac{\operatorname{Ar}(\triangle ABC )}{\operatorname{Ar}(\triangle ADB )}=\frac{ BC }{ AD } \\
= & \frac{4}{8} \\
= & \frac{1}{2}
\end{aligned}$
View full question & answer→Here, $\triangle ABC$ and $\triangle ADB$ has common Base.
$\therefore \frac{\operatorname{Ar}(\triangle ABC )}{\operatorname{Ar}(\triangle ADB )}=\frac{\text { height of } \triangle ABO }{\text { height of } \triangle ADB }$
(PROPERTY: Areas of triangles with equal bases are proportional to their corresponding heights.)
$\begin{aligned}
& \frac{\operatorname{Ar}(\triangle ABC )}{\operatorname{Ar}(\triangle ADB )}=\frac{ BC }{ AD } \\
= & \frac{4}{8} \\
= & \frac{1}{2}
\end{aligned}$
Question 612 Marks
Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.
Answer
View full question & answer→We know that area of triangle $=\frac{1}{2} \times$ Base $\times$ Height
$\begin{array}{l}
\Rightarrow \text { Area }(\text { triangle } 1)=\frac{1}{2} \times 9 \times 5 \\
=\frac{45}{2} \\
\Rightarrow \text { Area }(\text { triangle } 2)=\frac{1}{2} \times 10 \times 6 \\
=30
\end{array}$
Area(triangle 1 )
$\therefore$ the ratio of areas of these triangles will be $=\frac{\text { Area(triangle } 1)}{\text { Area(triangle } 2)}$
$\begin{array}{l}
=\frac{\frac{45}{2}}{30} \\
=\frac{45}{2} \times \frac{1}{30} \\
=\frac{3}{4}
\end{array}$
$\begin{array}{l}
\Rightarrow \text { Area }(\text { triangle } 1)=\frac{1}{2} \times 9 \times 5 \\
=\frac{45}{2} \\
\Rightarrow \text { Area }(\text { triangle } 2)=\frac{1}{2} \times 10 \times 6 \\
=30
\end{array}$
Area(triangle 1 )
$\therefore$ the ratio of areas of these triangles will be $=\frac{\text { Area(triangle } 1)}{\text { Area(triangle } 2)}$
$\begin{array}{l}
=\frac{\frac{45}{2}}{30} \\
=\frac{45}{2} \times \frac{1}{30} \\
=\frac{3}{4}
\end{array}$
Question 622 Marks
In figure 1.75, A – D – C and B – E – C seg DE || side AB If AD = 5, DC = 3, BC = 6.4 then find BE.
Answer
View full question & answer→By Basic Proportionality Theorem-
$\begin{array}{l}
\Rightarrow \frac{C D}{D A}=\frac{C E}{E B} \\
\Rightarrow \frac{3}{5}=\frac{6.4-x}{x} \\
\Rightarrow 3 x=32-5 x \\
\Rightarrow 8 x=32 \\
\Rightarrow x=4=B E
\end{array}$
$\begin{array}{l}
\Rightarrow \frac{C D}{D A}=\frac{C E}{E B} \\
\Rightarrow \frac{3}{5}=\frac{6.4-x}{x} \\
\Rightarrow 3 x=32-5 x \\
\Rightarrow 8 x=32 \\
\Rightarrow x=4=B E
\end{array}$
Question 632 Marks
Answer
View full question & answer→$\begin{array}{l}\text { In } \triangle APD \text { and } \triangle CPB \\ \Rightarrow \angle APD \cong \angle CPB \text { (opposite angles) } \\ \Rightarrow \angle ADP \cong \angle PBC \text { (Alternate angles } \because AD \| BC \text { ) } \\ \Rightarrow \triangle APD \sim \triangle CPB \text { (By AA Test) } \\ \Rightarrow \frac{ AP }{ PC }=\frac{ PD }{ BP } \text { (corresponding sides are proportional) } \\ \Rightarrow \frac{ AP }{ PD }=\frac{ PC }{ BP }\end{array}$
Question 642 Marks
$\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}, \mathrm{A}(\triangle \mathrm{ABC})=16, \mathrm{~A}(\triangle \mathrm{PQP})=25$, then find the value of ratio $\frac{\mathrm{AB}}{\mathrm{PQ}}$.
Answer
$\begin{array}{l} : \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} \\
\therefore \frac{\mathrm{A}(\triangle \mathrm{ABC})}{\mathrm{A}(\Delta \mathrm{PQR})}=\frac{\mathrm{AB}^2}{\mathrm{PQ}^2}\text{ .......... theorem of areas of similar triangles}\\
\therefore \frac{16}{25}=\frac{\mathrm{AB}^2}{\mathrm{PQ}^2} \quad \therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{4}{5}\text{.......... taking square roots}
\end{array}
$
theorem of areas of similar triangles taking square roots
View full question & answer→$\begin{array}{l} : \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} \\
\therefore \frac{\mathrm{A}(\triangle \mathrm{ABC})}{\mathrm{A}(\Delta \mathrm{PQR})}=\frac{\mathrm{AB}^2}{\mathrm{PQ}^2}\text{ .......... theorem of areas of similar triangles}\\
\therefore \frac{16}{25}=\frac{\mathrm{AB}^2}{\mathrm{PQ}^2} \quad \therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{4}{5}\text{.......... taking square roots}
\end{array}
$
theorem of areas of similar triangles taking square roots
Question 652 Marks
Ratio of corresponging sides of two similar triangles is 2:5, If the area of the small triangle is 64 sq.cm. then what is the area of the bigger triangle ?
Answer
View full question & answer→Assume that $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$.
$\triangle \mathrm{ABC}$ is smaller and $\triangle \mathrm{PQR}$ is bigger triangle.
\[
\begin{array}{l}
\therefore \frac{\mathrm{A}(\triangle \mathrm{ABC})}{\mathrm{A}(\triangle \mathrm{PQR})}=\frac{(2)^2}{(5)^2}=\frac{4}{25} \ldots \ldots . . . \text { ratio of areas of similar triangles } \\
\therefore \frac{64}{\mathrm{~A}(\Delta \mathrm{PQR})}=\frac{4}{25} \\
4 \times \mathrm{A}(\triangle \mathrm{PQR})=64 \times 25 \\
\mathrm{~A}(\Delta \mathrm{PQR})=\frac{64 \times 25}{4}=400 \\
\therefore \text { area of bigger triangle }=400 \mathrm{sq} . \mathrm{cm} .
\end{array}
\]
$\triangle \mathrm{ABC}$ is smaller and $\triangle \mathrm{PQR}$ is bigger triangle.
\[
\begin{array}{l}
\therefore \frac{\mathrm{A}(\triangle \mathrm{ABC})}{\mathrm{A}(\triangle \mathrm{PQR})}=\frac{(2)^2}{(5)^2}=\frac{4}{25} \ldots \ldots . . . \text { ratio of areas of similar triangles } \\
\therefore \frac{64}{\mathrm{~A}(\Delta \mathrm{PQR})}=\frac{4}{25} \\
4 \times \mathrm{A}(\triangle \mathrm{PQR})=64 \times 25 \\
\mathrm{~A}(\Delta \mathrm{PQR})=\frac{64 \times 25}{4}=400 \\
\therefore \text { area of bigger triangle }=400 \mathrm{sq} . \mathrm{cm} .
\end{array}
\]
Question 662 Marks
In trapezium $A B C D$, side $A B \|$ side $C D$, diagonal $A C$ and $B D$ intersect each other at point $P$. Then prove that $\frac{A(\Delta \mathrm{ABP})}{\mathrm{A}(\Delta \mathrm{CPD})}=\frac{\mathrm{AB}^2}{\mathrm{CD}^2}$.
Answer
View full question & answer→In trapezium $\mathrm{ABCD}$ side $\mathrm{AB} \|$ side $\mathrm{CD}$ In $\triangle \mathrm{APB}$ and $\triangle \mathrm{CPD}$
$\angle \mathrm{PAB} \cong \angle \mathrm{PCD}$ alternate angles
$\angle \mathrm{APB} \cong \angle \mathrm{CPD}$ opposite angles
$\therefore \triangle \mathrm{APB} \sim \triangle \mathrm{CPD}$ AA test of similarity
$\frac{\mathrm{A}(\Delta \mathrm{APB})}{\mathrm{A}(\Delta \mathrm{CPD})}=\frac{\mathrm{AB}^2}{\mathrm{CD}^2} \ldots \ldots$ theorem of areas of similar triangles
$\angle \mathrm{PAB} \cong \angle \mathrm{PCD}$ alternate angles
$\angle \mathrm{APB} \cong \angle \mathrm{CPD}$ opposite angles
$\therefore \triangle \mathrm{APB} \sim \triangle \mathrm{CPD}$ AA test of similarity
$\frac{\mathrm{A}(\Delta \mathrm{APB})}{\mathrm{A}(\Delta \mathrm{CPD})}=\frac{\mathrm{AB}^2}{\mathrm{CD}^2} \ldots \ldots$ theorem of areas of similar triangles
Question 672 Marks
In the adjoining figure BP ⊥ AC, CQ ⊥ AB, A – P- C, A- Q- B , then prove that △ APB and △ AQC are similar.
Answer
View full question & answer→In $\triangle A P B$ and $\triangle A Q C$,
$\angle A P B=90^{\circ}$$\quad$(I)
$\angle A Q C=90^{\circ}$$\quad$(II)
$\therefore \angle A P B=\angle A Q C$ from (I) and (II)
Also, $\angle A B P=\angle A C Q \quad$ (common acute angles)
$\therefore \triangle A P B \sim \triangle A Q C$ (by AA test)
$\triangle A P B \sim \triangle A Q C$
$\angle A P B=90^{\circ}$$\quad$(I)
$\angle A Q C=90^{\circ}$$\quad$(II)
$\therefore \angle A P B=\angle A Q C$ from (I) and (II)
Also, $\angle A B P=\angle A C Q \quad$ (common acute angles)
$\therefore \triangle A P B \sim \triangle A Q C$ (by AA test)
$\triangle A P B \sim \triangle A Q C$
Question 682 Marks
Can we say that the two triangles in figure 1.52 similar, according to information given? If yes, by which test ?
Answer
$\Delta \mathrm{XYZ}$ and $\Delta \mathrm{MNP}$,
$\begin{array}{l}
\frac{X Y}{M N}=\frac{14}{21}=\frac{2}{3}, \\
\frac{Y Z}{N P}=\frac{20}{30}=\frac{2}{3}
\end{array}$
It is given that $\angle \mathrm{Z} \cong \angle \mathrm{P}$.
But $\angle \mathrm{Z}$ and $\angle \mathrm{P}$ are not included angles by sides which are in proportion.
: $\Delta \mathrm{XYZ}$ and $\Delta \mathrm{MNP}$ can not be said to be similar.
View full question & answer→$\Delta \mathrm{XYZ}$ and $\Delta \mathrm{MNP}$,
$\begin{array}{l}
\frac{X Y}{M N}=\frac{14}{21}=\frac{2}{3}, \\
\frac{Y Z}{N P}=\frac{20}{30}=\frac{2}{3}
\end{array}$
It is given that $\angle \mathrm{Z} \cong \angle \mathrm{P}$.
But $\angle \mathrm{Z}$ and $\angle \mathrm{P}$ are not included angles by sides which are in proportion.
: $\Delta \mathrm{XYZ}$ and $\Delta \mathrm{MNP}$ can not be said to be similar.
Question 692 Marks
In $\triangle \mathrm{ABC}$, ray $\mathrm{BD}$ bisects $\angle \mathrm{ABC}$. $A-D-C$, side DE $\|$ side $B C, A-E-B$ then prove that, $\frac{A B}{B C}=\frac{A E}{E B}$
Answer
View full question & answer→In $\triangle A B C$, ray $B D$ bisects $\angle A B C$.
Points $A-D-C$ are collinear, $D E \| B C$, and points $A-E-B$ are collinear.
To Prove : $\frac{A B}{B C}=\frac{A E}{E B}$
Proof:
In $\triangle A B C$, ray $B D$ bisects $\angle A B C$.
$\therefore \frac{A B}{B C}=\frac{A D}{D C} \quad \ldots \ldots$
(I) (Angle bisector theorem)
In $\triangle A B C$, since $D E \| B C$,
$\frac{A E}{E B}=\frac{A D}{D C} \quad \ldots \ldots$
(II) (Basic proportionality theorem)
From equations (I) and (II),
$\frac{A B}{B C}=\frac{A E}{E B}$
Points $A-D-C$ are collinear, $D E \| B C$, and points $A-E-B$ are collinear.
To Prove : $\frac{A B}{B C}=\frac{A E}{E B}$
Proof:
In $\triangle A B C$, ray $B D$ bisects $\angle A B C$.
$\therefore \frac{A B}{B C}=\frac{A D}{D C} \quad \ldots \ldots$
(I) (Angle bisector theorem)
In $\triangle A B C$, since $D E \| B C$,
$\frac{A E}{E B}=\frac{A D}{D C} \quad \ldots \ldots$
(II) (Basic proportionality theorem)
From equations (I) and (II),
$\frac{A B}{B C}=\frac{A E}{E B}$
Question 702 Marks
In $\triangle PQR$, seg $RS$ bisects $\angle R$.
If $PR =15, RQ =20 PS =12$
then find $S Q$.
View full question & answer→If $PR =15, RQ =20 PS =12$
then find $S Q$.
Question 712 Marks
Given : line $l \|$ line $m \|$ line $n$ $t_1$ and $t_2$ are transversals. Transversal $t_1$ intersects the lines in points A, B, C and $\mathrm{t}_2$ intersects the lines in points $\mathrm{P}$, Q, R. To prove : $\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{PQ}}{\mathrm{QR}}$
View full question & answer→Question 722 Marks
$\triangle D E F \sim \triangle M N K$, If $D E=5$ and $M N=6$, then find the value of $A (\triangle DEF ): A (\triangle MNK )$
Answer
View full question & answer→$25: 36$
Question 732 Marks
The ratio of the areas of two triangles with equal height is $3: 2$. The base of the larger triangle is 18 cm . Find the corresponding base of the smaller triangle.
Answer
View full question & answer→12 cm
Question 742 Marks
Let X be any point on side BC of $\Delta$ABC. seg XM || side AB and seg XN || side CA. M-N-T, T-B-X. Prove that : TX2 = TB.TC.
Answer
View full question & answer→Since $X M \| A B$, in $\triangle A B C$, by the Basic Proportionality Theorem, we get
$\frac{B X}{X C}=\frac{B M}{M A}$ ....(1)
Since $X N \| C A$, in $\triangle A B C$, again by the Basic Proportionality Theorem, we get
$\frac{B X}{X C}=\frac{B N}{N A} \quad \ldots \ldots$ ....(2)
From equations (1) and (2),
$\frac{B M}{M A}=\frac{B N}{N A}$
Since points $M, N, T$ are collinear, and $T, B, X$ are collinear, by applying the Intercept Theorem, we get
$\frac{T B}{T X}=\frac{T X}{T C}$
Cross-multiplying, we obtain
$T X^2=T B \cdot T C$
$\frac{B X}{X C}=\frac{B M}{M A}$ ....(1)
Since $X N \| C A$, in $\triangle A B C$, again by the Basic Proportionality Theorem, we get
$\frac{B X}{X C}=\frac{B N}{N A} \quad \ldots \ldots$ ....(2)
From equations (1) and (2),
$\frac{B M}{M A}=\frac{B N}{N A}$
Since points $M, N, T$ are collinear, and $T, B, X$ are collinear, by applying the Intercept Theorem, we get
$\frac{T B}{T X}=\frac{T X}{T C}$
Cross-multiplying, we obtain
$T X^2=T B \cdot T C$
Question 752 Marks
In $\Delta$XYZ, XY = YZ. Ray YM bisects $\angle$XYZ. X-M-Z. Prove that M is midpoint of seg XZ.
Answer
View full question & answer→Point $M$ is the midpoint of segment $X Z$.
In $\triangle X Y Z$, it is given that
$X Y=Y Z$
Ray $Y M$ bisects $\angle X Y Z$.
$\therefore \angle X Y M=\angle M Y Z$
Now, in $\triangle X Y M$ and $\triangle M Y Z$,
1. $X Y=Y Z$ (Given)
2. $\angle X Y M=\angle M Y Z$ (Angle bisector)
3. $Y M=Y M \quad$ (Common side)
$\therefore \triangle X Y M \cong \triangle M Y Z \quad$ (by SAS congruence test)
$\therefore X M=M Z \quad$ (Corresponding parts of congruent triangles)
Since $M$ lies on segment $X Z$ and divides it into two equal parts,
$\therefore M$ is the midpoint of segment $X Z$.
In $\triangle X Y Z$, it is given that
$X Y=Y Z$
Ray $Y M$ bisects $\angle X Y Z$.
$\therefore \angle X Y M=\angle M Y Z$
Now, in $\triangle X Y M$ and $\triangle M Y Z$,
1. $X Y=Y Z$ (Given)
2. $\angle X Y M=\angle M Y Z$ (Angle bisector)
3. $Y M=Y M \quad$ (Common side)
$\therefore \triangle X Y M \cong \triangle M Y Z \quad$ (by SAS congruence test)
$\therefore X M=M Z \quad$ (Corresponding parts of congruent triangles)
Since $M$ lies on segment $X Z$ and divides it into two equal parts,
$\therefore M$ is the midpoint of segment $X Z$.
Question 762 Marks
In $\Delta$PQR, seg RS is bisector of $\angle P R Q$ PS = 6, SQ = 8, PR = 15. Find QR.
Answer
View full question & answer→ 20 units
Question 772 Marks
In $\Delta$DEF, line PQ || side EF. DQ = 1.8, QF = 5.4, PE = 7.2. Find DE.
Answer
View full question & answer→9.6 units
Question 782 Marks
If $\triangle ABC \sim \triangle DEF$ such that the area of $\triangle ABC$ is $9 cm^2$ and the area of $\triangle D E F$ is $16 cm^2$. If $B C=2.1$ cm . Find length of EF.
Answer
View full question & answer→2.8
Question 792 Marks
A vertical pole of a length 6 m casts a shadow of 4 m long on the ground. At the same time a tower casts a shadow 28 m long. Find the height of the tower.
Answer
View full question & answer→42 m