Download App

Maths 3 Marks Question

Triangles · Maharashtra Board STD 10 (MSBSHSE - English Medium). 52 questions.

Questions

3 Marks Question

Take a timed test

52 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.
Answer
Let the two triangles be ABC and PQR.
We have:
$\triangle\text{ABC}\sim\triangle\text{PQR}$
Here,
BC = a, AC = b and AB = c
PQ = r, PR = q and QR = p
We have to prove:
$\frac{\text{a}}{\text{p}}=\frac{\text{b}}{\text{q}}=\frac{\text{c}}{\text{r}}=\frac{\text{a+b+c}}{\text{p+q+r}}$
$\triangle\text{ABC}\sim\triangle\text{PQR};$ therefore, their corresponding sides will be proportional.
$\Rightarrow\frac{\text{a}}{\text{p}}=\frac{\text{b}}{\text{q}}=\frac{\text{c}}{\text{r}}=\text{k }(\text{say})\dots(\text{i})$
$\Rightarrow\text{a}=\text{kp},\text{b}=\text{kq}$ and $\text{c}=\text{kr}$
$\therefore\frac{\text{Perimeter of}\triangle\text{ABC}}{\text{Perimeter of}\triangle\text{PQR}}=\frac{\text{a+b+c}}{\text{p+q+r}}=\frac{\text{kp+kq+kr}}{\text{p+q+r}}=\text{k}\dots(\text{ii})$
From (i) and (ii), we get:
$\frac{\text{a}}{\text{p}}=\frac{\text{b}}{\text{q}}=\frac{\text{c}}{\text{r}}=\frac{\text{a+b+c}}{\text{p+q+r}}=\frac{\text{Perimeter of}\triangle\text{ABC}}{\text{Perimeter of}\triangle\text{PQR}}$
This completes the proof.
View full question & answer→
Question 23 Marks
The corresponding altitudes of two similar triangles are 6cm and 9cm respectively, Find the ratio of their areas.
Answer
Let the two triangle be ABC and DEF with altitudes AP and DQ, respectively.

It is given that $\triangle\text{ABC}\sim\triangle\text{DEF}.$
We know that the ratio of areas of two similar triangle is equia to the ratio of squares of their corresponding altitudes.
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{(\text{AP})^2}{(\text{DQ})^2}$
$\Rightarrow\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{6}^2}{\text{9}^2}$
$=\frac{36}{81}$
$=\frac{4}{9}$
Hence, the ratio of their areas is 4 : 9
View full question & answer→
Question 33 Marks
In $\triangle\text{ABC},\text{AB}=\text{AC}.$ Side BC is produced to D. prove that
$(\text{AD}^2-\text{AC}^2)=\text{BD}.\text{CD}$
Answer


Constructing: Draw $\text{AE}\perp\text{BC}.$
Since $\triangle\text{ABC}$ is an isosceles triangle.
We know that in an isosceles triangle,
the altitude and median are the same.
So, $CE = BE$
$\Rightarrow DE + CE = BE + DE = BD$
In $\triangle\text{AED},$
By pythagoras Theorem,
$AD^2 = AE^2 + DE^2$
$\Rightarrow AE^2 = AD^2 - DE^2 ......(i)$
In $\triangle\text{AEC},$
By Pythagoras Theorem,
$AC^2 = AE^2 + EC^2$
$\Rightarrow AE^2 = AC^2 - EC^2 ....(ii)$
From (i) and (ii), we have
$AD^2 - DE^2 = AC - EC^2$
$\Rightarrow AD^2 - AC^2 = DE^2 - EC^2$
$\Rightarrow AD^2 - AC^2 = (DE - EC) (DE + EC)$
$\Rightarrow AD^2 - AC^2 = (CD) (BD)$
$\Rightarrow AD^2 - AC^2 = BD \times CD$
Hence proved.
View full question & answer→
Question 43 Marks
D and E are points on the sides AB and AC respectively of a $\triangle\text{ABC}$ such that DE || BC:
AD = (7x - 4)cm, AE = (5x - 2)cm, DB = (3x + 4)cm and EC = 3x cm.
Answer
In $\triangle\text{ABC},$ it is given that DE || BC.
Applying Thales' theorem, we get:
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{7\text{x}-4}{3\text{x}+4}=\frac{5\text{x}-2}{3\text{x}}$
$\Rightarrow3\text{x}(7\text{x}-4)=(5\text{x}-2)(3\text{x}+4)$
$\Rightarrow21\text{x}^2-12\text{x}=15\text{x}^2+14\text{x}-8$
$\Rightarrow6\text{x}^2-26\text{x}+8=0$
$\Rightarrow\big(\text{x}-4\big)\big(6\text{x}-2\big)=0$
$\Rightarrow\text{x}=4,\frac{1}{3}$
$\therefore\text{x}\not=\frac{1}{3}$ (as if $\text{x}=\frac{1}{3}$ then AE will become negative)
$\therefore\text{x}=4\text{cm}$
View full question & answer→
Question 53 Marks
D and E are points on the sides AB and AC respectively of a $\triangle\text{ABC}.$ In the following cases, determine whether DE || BC or not. AD = 7.2cm, AE = 6.4cm, AB = 12cm and AC = 10cm.
Answer
We have:
AD = 7.2cm, AB = 12cm
Therefore,
DB = 12 - 7.2 = 4.8cm
Similarly,
AE = 6.4cm, AC = 10cm
Therefore,
EC = 10 - 6.4 = 3.6cm
Now,
$\frac{\text{AD}}{\text{DB}}=\frac{7.2}{4.8}=\frac{3}{2}$
$\frac{\text{AE}}{\text{EC}}=\frac{6.4}{3.6}=\frac{16}{9}$
Thus, $\frac{\text{AD}}{\text{DB}}\not=\frac{\text{AE}}{\text{EC}}$
Applying the converse of Thalse' theorem, we conclude thet DE is not parallel to BC.
View full question & answer→
Question 63 Marks
$\triangle\text{ABC}$ is an isosceles triangle with $AB = AC = 13\ cm$. The length of altitude from $A$ on $BC$ is $5\ cm$. Find $BC$.
Answer
Given: $\triangle\text{ABC}$ is an isosceles triangle with $AB = AC = 13\ cm^2$ Const:
Draw altitude from A to BC $(\text{AL}\perp\text{BC} ).$
Now, $AL = 5\ cm$​​​​​​​

​​​​​​​
In $\triangle\text{ALB},$ $\angle\text{ALB}=90^\circ$
$\therefore\text{AB}^2=\text{AL}^2+\text{BL}^2$(by pythagoras theorem)
$\therefore13^2=(5)^2+\text{BL}^2$ $(169-25)\text{cm}^2=\text{BL}^2$
$\text{BL}^2=144\text{cm}^2$ $\text{BL}=\sqrt{144}\text{cm}=12\text{cm}$ In $\triangle\text{ALC},$
$\angle\text{ALC}=90^\circ$ $\text{AC}^2=\text{AL}^2+\text{LC}^2$
$\Rightarrow\text{LC}^2=\Big(\text{AC}^2-\text{AL}^2\Big)$
$=\Big[(13)^2-(5)^2\Big]\text{cm}^2$ $=(169-25)\text{cm}^2$
$=144\text{cm}^2$
$=\sqrt{144}=12\text{cm}$ $\therefore\text{BC}=\text{BL}+\text{LC}$
$=(12+12)\text{cm}=24\text{cm}$
View full question & answer→
Question 73 Marks
The sides of certain triangles are given below. Determine them are right triangles:
9cm, 16cm, 18cm.
Answer
For a given triangle to be a right angled, the sum of the squares of the two sides must be equal to the square of the largest side.
Let a = 9cm, b = 16cm and c= 18cm. Then
$\Big(\text{a}^2+\text{b}^2\Big)=\Big[9^2+(16)^2\Big]$
$=(81+256)\text{cm}^2$
$=337\text{cm}^2$
and $\text{c}^2=(18)^2\text{cm}^2=324\text{cm}^2$
$\therefore\big(\text{a}^2+\text{b}^2\big)\not=\text{c}^2$
Hence the given triangle is not right angled.
View full question & answer→
Question 83 Marks
In the given figure, $DE || BC$. If $DE = 3\ cm$, $BC = 6\ cm$ and $\text{ar}(\triangle\text{ADE})=15\text{cm}^2,$ find the area of $\triangle\text{ABC}.$

Answer
It is given that $DE || BC$
$\therefore\angle\text{ADE}=\angle\text{ABC}$ (Corresponding angles)
$\angle\text{AED}=\angle\text{ACB}$ (Corresponding angles)
By AA similarity, we can conclude that $\triangle\text{ADE}\sim\triangle\text{ABC}.$
$\therefore\frac{\text{ar}(\triangle\text{ADE})}{\text{ar}(\triangle\text{ABC})}=\frac{\text{DE}^2}{\text{BC}^2}$
$\Rightarrow\frac{15}{\text{ar}(\triangle\text{ABC})}=\frac{3^2}{6^2}$
$\Rightarrow\text{ar}(\triangle\text{ABC})=\frac{15\times36}{9}$
$=60\text{cm}^2$
Hence, area of triangle ABC is $60cm^2$
View full question & answer→
Question 93 Marks
In a trapezium ABCD, it is given that AB || CD and AB = 2CD. Its diagonals AC and BD intersect at the point O such that $\text{ar}(\triangle\text{AOB})=84\text{cm}^2.$ Find $\text{ar}(\triangle\text{COD}).$
Answer
In $\triangle\text{AOB}$ and $\triangle\text{COD},$ we have:

$\angle\text{AOB}=\angle\text{COD}$ (Vertically opposite angles)
$\angle\text{OAB}=\angle\text{OCD}$ (Alternate angles as AB || CD)
Applying AA similiarity criterion, we get:
$\triangle\text{AOB}\sim\triangle\text{COD}$
$\therefore\frac{\text{ar}(\triangle\text{AOB})}{\text{ar}(\triangle\text{COD})}=\frac{\text{AB}^2}{\text{CD}^2}$
$\Rightarrow\frac{84}{\text{ar}(\triangle\text{COD})}=\Big(\frac{\text{AB}}{\text{CD}}\Big)^2$
$\Rightarrow\frac{84}{\text{ar}(\triangle\text{COD})}=\Big(\frac{2\text{CD}}{\text{CD}}\Big)^2$
$\Rightarrow\text{ar}(\triangle\text{COD})=\frac{84}{4}=21\text{cm}^2$
View full question & answer→
Question 103 Marks
In the given figure, $\angle\text{AMN}=\angle\text{MBC}=76^\circ.$ If p, q and r are the lengths of AM, MB and BC respectively then express the length of MN in terms of p, q and r.
Answer
In $\triangle\text{ABC}$ and $\triangle\text{AMN},$
$\angle\text{M}=\angle\text{B}=76^\circ$ .....(corresponding angles)
and $\angle\text{BAC}=\angle\text{MAN}$ .....(common angle)
$\Rightarrow\triangle\text{ABC}\sim\triangle\text{AMN}$ .....(AA criterion)
$\frac{\text{AB}}{\text{AM}}=\frac{\text{BC}}{\text{MN}}$
$\frac{\text{AM}+\text{MB}}{\text{AM}}=\frac{\text{BC}}{\text{MN}}$
$\frac{\text{a}+\text{b}}{\text{a}}=\frac{\text{c}}{\text{MN}}$
$\Rightarrow\text{MN}=\frac{\text{ac}}{\text{a}+\text{b}}$
View full question & answer→
Question 113 Marks
Find the height of an equilateral triangle of side 12cm.
Answer
$\triangle\text{ABC}$ is an equilateral triangle in which all side are equal. Therefore, AB = BC = AC = 12cm If BC = 12cm Then, BD = BC = 6cm
In $\triangle\text{ADB},$ $\text{AB}^2=\text{AD}^2+\text{BD}^2$ (By applying pythagoras theorem) $\text{AD}^2=\text{AB}^2-\text{BD}^2$ $\text{AD}^2=\Big[(12)^2-(6)^2\Big]\text{cm}^2$ $\text{AD}^2=\sqrt{108}\text{cm}$ $\text{AD}=\sqrt{108}\text{cm}=6\sqrt{3}\text{cm}$ Hence the height of the triangle is $6\sqrt{3}\text{cm}$
View full question & answer→
Question 123 Marks
$\triangle\text{ABC}\sim\triangle\text{PQR}$ and $\text{ar}(\triangle\text{ABC})=4\text{ar}(\triangle\text{PQR}).$ If BC 12cm, find QR.
Answer
Given: $\text{ar}(\triangle\text{ABC})=4\text{ar}(\triangle\text{PQR})$
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\frac{\text{4}}{\text{1}}$
$\therefore\triangle\text{ABC}\sim\triangle\text{PQR}$
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\frac{\text{BC}^2}{\text{QR}^2}$
$\therefore\frac{\text{BC}^2}{\text{QR}^2}=\frac{\text{4}}{\text{1}}$
$\Rightarrow\text{QR}^2=\frac{12^2}{4}$
$\Rightarrow\text{QR}^2=36$
$\Rightarrow\text{QR}=6$
Hence, QR = 6cm
View full question & answer→
Question 133 Marks
$\triangle\text{ABC}\sim\triangle\text{DEF}$ such that $\text{ar}(\triangle\text{ABC})=64\text{cm}^2$ and $\text{ar}(\triangle\text{DEF})=169\text{cm}^2.$ If BC = 4cm, find EF.
Answer
$\therefore\triangle\text{ABC}\sim\triangle\text{DEF}$
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{BC}^2}{\text{EF}^2}$
$\Rightarrow\frac{64}{169}=\frac{4^2}{\text{EF}^2}$
$\Rightarrow\text{EF}^2=\frac{16\times169}{\text{64}}$
$\Rightarrow\text{EF}=\frac{4\times13}{\text{8}}=6.5\text{cm}$
View full question & answer→
Question 143 Marks
The corresponding sides of two similar triangles are in the ratio $2 : 3$. If the area of the smaller triangle is $48\ cm^2$​​​​​​​, find the area of the larger triangle.
Answer
It is given that the triangles are similar.
Therefore, the ratio of areas of similar triangles will be equal to the ratio of squares of their corresponding sides.
$\therefore\frac{48}{\text{Area of larger triangle}}=\frac{2^2}{3^2}$
$\Rightarrow\frac{48}{\text{Area of larger triangle}}=\frac{4}{9}$
$\Rightarrow\text{Area of larger triangle}=\frac{48\times9}{4}=108\text{cm}^2$
View full question & answer→
Question 153 Marks
ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that $\text{CQ}=\frac{1}{4}\text{AC}.$ If PQ produced meets BC at R, prove that R is the midpoint of BC.
Answer
We know that the diagonals of a parallelogram bisect each other.
Therefore,
$\text{CS}=\frac{1}{2}\text{AC}\dots(\text{i})$
Also, it is given that $\text{CQ}=\frac{1}{4}\text{AC}\dots(\text{ii})$
Dividing equation (ii) by (i), we get:
$\frac{\text{CQ}}{\text{CS}}=\frac{\frac{1}{4}\text{AC}}{\frac{1}{2}\text{AC}}$
or, $\text{CQ}=\frac{1}{2}\text{CS}$
Hence, Q is the midpoint of CS.
Therefore, according to midpoint theorem in $\triangle\text{CSD}$
PQ || DS
If PQ || DS, we can say that QR || SB
In $\triangle\text{CSB},\text{Q}$ is midpoint of CS and QR || SB.
Applying converse of midpoint theorem, we conclude that R the midpoint of CB.
This comletes the proof .
View full question & answer→
Question 163 Marks
In the given figure, $\triangle\text{ODC}\sim\triangle\text{OBA},\angle\text{BOC}=115^\circ$ and $\angle\text{CDO}=70^\circ.$
Find
  1. $\angle\text{DOC}$
  2. $\angle\text{DCO}$
  3. $\angle\text{OAB}$
  4. $\angle\text{OBA}$
Answer

$\triangle\text{ODC}\sim\triangle\text{OBC}$
$\angle\text{BOC}=115^\circ$
$\angle\text{CDO}=70^\circ$
  1. $\angle\text{DOC}=(180^\circ-\angle\text{BOC})$
$=(180^\circ-115^\circ)$

$=65^\circ$
  1. $\angle\text{OCD}=180^\circ-\angle\text{CDO}-\angle\text{DOC}$
$\angle\text{OCD}=180^\circ-(70^\circ+65^\circ)$

$=45^\circ$
  1. Now, $\triangle\text{ABO}\sim\triangle\text{ODC}$
$\angle\text{AOB}=\angle\text{OCD }(\text{vert.Opp }\angle\text{s})=45^\circ$

$\angle\text{OAB}=\angle\text{OCD}=45^\circ$
  1. $\angle\text{OBA}=\angle\text{ODC}$ (alternate angles) = 70º
So, $\angle\text{OAB}=45^\circ$ and $\angle\text{OBA}=70^\circ$
View full question & answer→
Question 173 Marks
The corresponding sides of two similar triangles ABC and DEF are BC = 9.1cm and EF = 6.5cm. If the rerimeter of $\triangle\text{DEF}$ is 25cm, find the perimeter of $\triangle\text{ABC}.$
Answer
$\triangle\text{ABC}$ and $\triangle\text{DEF}$ are two similar triangles, therefore corresponding sides of both the triangle are proportional.
Hence, $\frac{\text{Perimeter of }\triangle\text{ABC}}{\text{Perimeter of }\triangle\text{DEF}}=\frac{\text{BC}}{\text{EF}}$
Let peremeter of $\triangle\text{ABC}=\text{x} \text{ cm}$
$\therefore\frac{\text{x}}{25}=\frac{9.1}{6.5}$
$\text{x}=\frac{9.1\times25}{6.5}=35\text{cm}$
Hence, perimeter of $\triangle\text{ABC}=35\text{ cm}$
View full question & answer→
Question 183 Marks
Two triangles DEF and GHK are such that $\angle\text{D}=48^\circ$ and $\angle\text{H}=57^\circ.$ If $\triangle\text{DEF}\sim\triangle\text{GHK}$ then find the measure of $\angle\text{F}.$
Answer
Given that $\triangle\text{DEF}\sim\triangle\text{GHK}.$
$\angle\text{D}=\angle\text{G}=48^\circ\dots(\text{Given})$
$\angle\text{E}=\angle\text{H}=57^\circ\dots(\text{Given})$
In $\triangle\text{DEF},$
$\angle\text{D}+\angle\text{E}+\angle\text{F}=180^\circ\dots(\text{Angel Sum Property})$
$\Rightarrow48^\circ+57^\circ+\angle\text{F}=180^\circ$
$\Rightarrow\angle\text{F}=75^\circ$
View full question & answer→
Question 193 Marks
For the following statments state whether true (T) or false(F):
The ratio of the areas of two similar triangles is equal to the ratio of their corresponding angle-bisector segments.
Answer
True.
Solution:

Given that $\triangle\text{ABC}\sim\triangle\text{DEF}$ in which AX and DY are the bisectors of $\angle\text{A}$ and $\angle\text{D}$ respectively.
To prove: $\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{AX}^2}{\text{DY}^2}$
We know that,
The ratio of the areas of two similar triangles is equal to the ratio of the squares of the squares of the squares of the corresponding sides.
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{AB}^2}{\text{DE}^2}\dots\text{(i)}$
$\triangle\text{ABC}\sim\triangle\text{DEF}$
$\Rightarrow\angle\text{A}=\angle\text{D}$
$\Rightarrow\frac{1}{2}\angle\text{A}=\frac{1}{2}\angle\text{D}$
$\Rightarrow\angle\text{BAX}=\frac{1}{2}\angle\text{EDY}$
In $\triangle\text{ABX}$ and $\triangle\text{DEY},$
$\angle\text{BAX}=\angle\text{EDY}$ and $\angle\text{B}=\angle\text{E}$
So, by the AA criterion for similarity,
$\triangle\text{ABX}\sim\triangle\text{DEY}$
$\Rightarrow\frac{\text{AB}}{\text{DE}}=\frac{\text{AX}}{\text{DY}}$
$\Rightarrow\frac{\text{AB}^2}{\text{DE}^2}=\frac{\text{AX}^2}{\text{DY}^2}\dots(\text{ii})$
From (i) and (ii),
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{AX}^2}{\text{DY}^2}$
View full question & answer→
Question 203 Marks
D and E are points on the sides AB and AC respectively of a $\triangle\text{ABC}.$ In the following cases, determine whether DE || BC or not.
AB = 10.8cm, AD = 6.3cm, AC = 9.6cm and EC = 4cm.
Answer
We have:
AB = 10.8cm, AD = 6.3cm
Therefore,
DB = 10.8 - 6.3 = 4.5cm
Similarly,
AC = 9.6cm, EC = 4cm
Therefore,
AE = 9.6 - 4 = 5.6cm
Now,
$\frac{\text{AD}}{\text{DB}}=\frac{6.3}{4.5}=\frac{7}{5}$
$\frac{\text{AE}}{\text{EC}}=\frac{5.6}{4}=\frac{7}{5}$
$\Rightarrow\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
Applying the converse of Thalse' theorem, we conclude thet DE || BC.
View full question & answer→
Question 213 Marks
State the midpoint theorem.
Answer
The line segment joining the mid-points of any two sides of a triangle is parallel to the third side.
View full question & answer→
Question 223 Marks
A guy wire attached to a vertical pole of height $18\ m$ is $24\ m$ long and has a stake attached to the other end.
How far from the base of the pole should the stake be driven so that the wire will be taut?
Answer



Let AC be a guy wire attached to be a pole AB of height $18\ m$.
Let BC be the distance that the stake is away from the pole so that the wire will be taut.
In $\triangle\text{ABC},$
By Pythagoras Theorem,
$AC^2 = AB^2 + BC^2$
$\Rightarrow 24^2 = 18^2 + BC^2$
$\Rightarrow BC^2 = 24^2 - 18^2$
$\Rightarrow BC^2 = 576 - 324$
$\Rightarrow BC^2 = 252$
$\Rightarrow\text{BC}^2=6\sqrt{7}\text{m}$
So, the stake should be $6\sqrt{7}\text{m}$ away from the pole so that the wire will be taut.
View full question & answer→
Question 233 Marks
In the given figure, D is the midpoint of side BC and $\text{AE}\perp\text{BC}.$ If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that.
$(\text{b}^2-\text{c}^2)=2\text{ax}$
Answer
Given: D is the midpoint of side $\text{BC},\text{AE}\perp\text{BC},\text{BC}=\text{a},\text{AC}=\text{b},\text{AB}=\text{c},\text{ED}=\text{x},\text{AD}=\text{p}$ and $\text{AE}=\text{h}$
In $\triangle\text{AEC},\angle\text{AEC}=90^\circ$
$\text{AD}^2=\text{AE}^2+\text{ED}^2$ (by pythagoras theorem)
$\Rightarrow\text{p}^2=\text{h}^2+\text{x}^2$
(iv) Subtracting (2) from (1), we get
$\text{b}^2-\text{c}^2=\text{p}^2+\text{ax}+\frac{\text{a}^2}{4}-\bigg(\text{p}^2-\text{ax}+\frac{\text{a}^2}{4}\bigg)$
$=\text{p}^2+\text{ax}+\frac{\text{a}^2}{4}-\text{p}^2+\text{ax}-\frac{\text{a}^2}{4}$
$\big(\text{b}^2-\text{c}^2\big)=2\text{ax}$
View full question & answer→
Question 243 Marks
The sides of certain triangles are given below. Determine them are right triangles:
1.4cm, 4.8cm, 5cm.
Answer
For a given triangle to be a right angled, the sum of the squares of the two sides must be equal to the square of the largest side.
Let a = 1.4cm, b = 4.8cm, and c = 5cm
$\big(\text{a}^2+\text{b}^2\big)=\big[(1.4)^2+(4.8)^2\big]\text{cm}^2$
$=(1.96+23.04)\text{cm}^2=25\text{cm}^2$
$\text{c}^2=(5\text{cm)}^2=25\text{cm}^2$
$\therefore\big(\text{a}^2+\text{b}^{2}\big)=\text{c}^2$
Hence, the given triangle is a right triangle.
View full question & answer→
Question 253 Marks
An aeroplane leaves an airport and flies due north at a speed of $1000\  km$ per hour. At the same time, aeroplane leaves the same airport and flies due west at a speed of $1200\ km$ per hour. How far apart will be the two planes after $1\frac{1}{2}\text{hour}?$
Answer


Let $C$ be the airport and let $A$ and $B$ be the two aerplanes flying north and west respectively Distance covered by $A$ in $1\frac{1}{2}=\frac{3}{2}\text{hours}$
$= AC$
$=1000\times\frac{3}{2}=1500\text{km}$
Distance covered by B in $1\frac{1}{2}=\frac{3}{2}\text{hours}$ hours
$= BC$
$=1200\times\frac{3}{2}=1800\text{km}$
In $\triangle\text{ABC},$
By Pythagoras theorem,
$AB^2 = AC^2 + BC^2$
$\Rightarrow AB^2= 1500^2 + 1800^2$
$\Rightarrow AB^2 = 2250000 + 3420000$
$\Rightarrow AB^2 = 5490000$
$\Rightarrow\text{AB}=300\sqrt{61}\text{km}$
Hence, after $1\frac{1}{2}$ hours, the planes $ A$ and $B$ are $300\sqrt{61}\text{km}$ apart.
View full question & answer→
Question 263 Marks
For the following statments state whether true (T) or false(F):
Any two rectangles are similar.
Answer
False.Solution:
Two rectangles are similar if their cirresponding sides are proportional.
View full question & answer→
Question 273 Marks
The sides of certain triangles are given below. Determine them are right triangles:
1.6cm, 3.8cm, 4cm.
Answer
For a given triangle to be a right angled, the sum of the squares of the two sides must be equal to the square of the largest side.
let a = 1.6cm, b = 3.8cm and c = 4cm
$\big(\text{a}^2+\text{b}^2\big)=\big[(1.6)^2+{(3.8})^2\big]\text{cm}^2$
$=(2.56+14.44)\text{cm}^2=17\text{cm}^2$
$\text{c}^2=(4)^2=16\text{cm}^2$
$\therefore\big(\text{a}^2+\text{b}^2\big)\not=\text{c}^2$
Hence, the given triangle is a right triangle.
View full question & answer→
Question 283 Marks
D and E are points on the sides AB and AC respectively of a $\triangle\text{ABC}$ such that DE || BC:
If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC.
Answer
In $\triangle\text{ABC},$ it is given that DE || BC.
Applying Thales' theorem, we get:
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\therefore\text{AD}=3.6,\text{AB}=10\text{cm},\text{AE}=4.5\text{cm}$
$\therefore\text{DB}=10-3.6=6.4\text{cm}$
or, $\frac{3.6}{6.4}=\frac{4.5}{\text{EC}}$
or, $\text{EC}=\frac{6.4\times4.5}{3.6}$
or, $\text{EC}=8\text{cm}$
Thus, $\text{AC}=\text{AE}+\text{EC}$
$=4.5 +8=12.5\text{cm}$
View full question & answer→
Question 293 Marks
State Pythagoras theoram.
Answer
In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
View full question & answer→
Question 303 Marks
In a $\triangle\text{ABC},\text{AD}$ is the bisector of $\angle\text{A}.$ If AB = 5.6cm, BD = 3.2cm and BC = 6cm, find AC.
Answer
It is given that AD bisects $\angle\text{A}.$
Applying angle-bisector theorem in $\triangle\text{ABC},$ we get:
$\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}}{\text{AC}}$
BD = 3.2cm, BC = 6cm
Therefore, DC = 6 - 3.2 = 2.8cm
$\Rightarrow\frac{3.2}{2.8}=\frac{5.6}{\text{AC}}$
$\Rightarrow\text{AC}=\frac{5.6\times2.8}{3.2}=4.9\text{cm}$
View full question & answer→
Question 313 Marks
State the two properties which are necessary for given two triangles to be similar.
Answer
Two triangles are said to be similar to each other if:
  1. Their corresponding angles are equal.
  2. Their corresponding sides are proportional.
View full question & answer→
Question 323 Marks
A man goes $10\ m$ due south and then $24\ m$ due west. How far is he from the starting point?
Answer
starting from O, let the man goas from O to A and then A to B as shows in the figure.
Then, $\text{OA}=10\text{m},\text{AB}=24\text{m }$ and $\angle\text{OAB}=90^\circ$




Using Pythagoras theorem:
$OB^2 = OA^2+AB^{2}$
$ \Rightarrow OB^2= 10^2 + 24^{2}$
$​​​​​​​ \Rightarrow OB^2 = 100 + 576$
$\Rightarrow OB^2 = 676$
$\Rightarrow\text{OB}=\sqrt{676}=26\text{m}$ Hence, the man is $26\ m$ south-west from the starting position.
View full question & answer→
Question 333 Marks
In the given figure, side BC of $\triangle\text{ABC}$ is bisected at D and O is any point on AD. BO and CO produced meet AC and AB at E and F respectively, and AD is produced to X so that D is the midpoint of OX Prove that AO : AX = AF : AB and show that EF || BC.
Answer
It is given that BC is bisected at D.
$\therefore\text{BD}=\text{DC}$
It is also given that OD = OX
The diagonals OX and BC of quadrilateral BOCX bisect each other.
Therefore, BOCX is a parallelogram
$\therefore\text{BO }||\text{ CX}$ and $\text{BX }||\text{ CO}$
$\text{BX }||\text{ CF}$ and $\text{CX }||\text{ BE}$
$\text{BX }||\text{ OF}$ and $\text{CX }||\text{ OE}$
Applying Thales' theorem in $\triangle\text{ABX},$ we get:
$\frac{\text{AO}}{\text{AX}}=\frac{\text{AF}}{\text{AB}}\dots(1)$
Also, in $\triangle\text{ACX},\text{CX }||\text{ OE}.$
Therefore by Thales' theorem, we get:
$\frac{\text{AO}}{\text{AX}}=\frac{\text{AE}}{\text{AC}}\dots(2)$
From (1) and (2), we have:
$\frac{\text{AF}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}$
Applying the converse of Thales' theorem in $\triangle\text{ABC},\text{EF }||\text{ CB}.$
This completes the proof.
View full question & answer→
Question 343 Marks
A 13-m-long ladder reaches a window of a building 12m above the ground. Determine the distance of the foot of the ladder from the building.
Answer
Let AB be the building and CB be the ladder. Then, $\text{AB}=12\text{m},\text{CB}=13\text{m}$ and $\angle\text{CAB}=90^\circ$
By Pythagoaras theoram, we have $\text{CB}^2=\text{AB}^2+\text{AC}^2$ $\text{AC}^2=\big[\text{CB}^2-\text{AB}^2\big]$ $=\Big[(13)^2-(12)^2\Big]\text{m}^2$ $=(169-144)\text{m}^2$ $=25\text{m}^2$ $\Rightarrow\text{AC}=\sqrt{25}\text{m}=5\text{m}$ Hence, the distance of the fool of the ladder from the building is 5m.
View full question & answer→
Question 353 Marks
D and E are points on the sides AB and AC respectively of a $\triangle\text{ABC}$ such that DE || BC: If $\frac{\text{AD}}{\text{AB}}=\frac{8}{15}$ and EC = 3.5cm, find AE.
Answer
In $\triangle\text{ABC},$ it is given that DE || BC.
Applying Thales' theorem, we get:
$\frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}$
$\Rightarrow\frac{\text{8}}{\text{15}}=\frac{\text{AE}}{\text{AE}+\text{EC}}$
$\Rightarrow\frac{\text{8}}{\text{15}}=\frac{\text{AE}}{\text{AE}+\text{3.5}}$
$\Rightarrow8\text{AE}+28=15\text{AE}$
$\Rightarrow7\text{AE}=28$
$\Rightarrow\text{AE}=4\text{cm}$
View full question & answer→
Question 363 Marks
In a $\triangle\text{ABC},\text{M}$ and N are points on the sides AB and AC respectively such that BM || BC.
Answer

In $\triangle\text{ABC},\angle\text{B}=\angle\text{C}$
$\therefore\text{AB}=\text{AC}$ (Sides opposite to equal angle are equal)
Subtracting BM from both sides, we get
AB - BM = AC - BM
⇒ AB - BM = AC - CN $(\therefore\text{BM=CN})$
⇒ AM = AN
$\therefore\angle\text{AMN}=\angle\text{ANM}$ (Angels opposite to equal sides are equal)
Now, in $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ\dots(1)$(Angle Sum Property of triangle)
Again in $\triangle\text{AMN},$
$\angle\text{A}+\angle\text{AMN}+\angle\text{ANM}=180^\circ\dots(2)$(Angle Sum Property of triangle)
From (1) and (2), we get
$\angle\text{B}+\angle\text{C}=\angle\text{AMN}+\angle\text{ANM}$
$\Rightarrow2\angle\text{B}=2\angle\text{ANM}$
$\Rightarrow\angle\text{B}=\angle\text{AMN}$
Since, $\angle\text{B}$ and $\angle\text{AMN}$ are corresponding angles.
$\therefore\text{MN }||\text{ BC}$
View full question & answer→
Question 373 Marks
In $\triangle\text{ABC},\text{D}$ is the midpoint of $BC$ and $\text{AE}\perp\text{BC}.$ If $\text{AC}>\text{AB},$ show that.
$\text{AB}^2=\text{AD}^2-\text{BC}.\text{DE}+\frac{1}{4}\text{BC}^2.$
Answer
Given: $\triangle\text{ABC}$ in which $D$ is the midpoint of $BC$. $\text{AE}\perp\text{BC}$ and $AC > AB.$ Then $BD = CD$ and $\angle\text{AED}=90^\circ,$ Then, $\angle\text{ADE}<90^\circ$ and $\angle\text{ADC}>90^\circ$
In $\triangle\text{AED},$ $\angle\text{AED}=90^\circ$ $\therefore\text{AD}^2=\text{AE}^2+\text{DE}^2$ $\Rightarrow\text{AE}^2=\big(\text{AD}^2-\text{DE}^2\big)\dots(1)$ In $\triangle\text{AEB},\angle\text{AEB}=90^\circ$ $\therefore \text{AB}^2=\text{AE}^2+\text{BE}^2\dots(2)$ Putting value of $AE^2 $ from (1) in (2), we get $\therefore\text{AB}^2=\big(\text{AD}^2-\text{DE}^2\big)+\text{BE}^2$ $=\big(\text{AD}^2-\text{DE}^2\big)+\big(\text{BD}^2-\text{DE}^2\big)\Big[\text{But}\text{ BD}=\frac{1}{2}\text{BC}\Big]$ $=\text{AD}^2-\text{DE}^2+\Big(\frac{1}{2}\text{BC}-\text{DE}\Big)^2$ $=\text{AD}^2-\text{DE}^2+\frac{1}{4}\text{BC}^2+\text{DE}^2-\text{BC.DE}$ $\text{AB}^2=\text{AD}^2-\text{BC},\text{DE}+\frac{1}{4}\text{BC}^2$
View full question & answer→
Question 383 Marks
D and E are points on the sides AB and AC respectively of a $\triangle\text{ABC}.$ In the following cases, determine whether DE || BC or not.
AD = 5.7cm, DB = 9.5cm, AE = 4.8cm and EC = 8cm.
Answer
We have:
$\frac{\text{AD}}{\text{DB}}=\frac{5.7}{9.5}=0.6\text{cm}$
$\frac{\text{AE}}{\text{EC}}=\frac{4.8}{8}=0.6\text{cm}$
Hence, $\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
Applying the converse of Thalse' theorem, we conclude thet DE || BC.
View full question & answer→
Question 393 Marks
ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the midpoints of AB, AC, CD and BD respectively, show that PQRS is a rhombus.
Answer

Given: ABCD is a quadrilateral in which AD = BC. P, Q, R, S are the midpoint of AB, AC, CD and BD.
To prove: PQRS is a rhombus
Proof: In $\triangle\text{ABC},$
Since P and Q are mid point of AB and AC
Therefore,PQ || BC and $\text{PQ}=\frac{1}{2}\text{BC}=\frac{1}{2}\text{DA}$ (Mid-point theorem)
Similarly,
In $\triangle\text{CDA},$
Since R and Q are mid point of CD and AC
Therefore, RQ || DA and $\text{PQ}=\frac{1}{2}\text{DA}$
In $\triangle\text{BDA},$
Since S and P are mid point of BD and AB
Therefore, SP || DA and $\text{SP}=\frac{1}{2}\text{DA}$
In $\triangle\text{CDB},$
Since S and P are mid point of BD and CD
Therefore, SR || BC and $\text{SR}=\frac{1}{2}\text{BC}=\frac{1}{2}\text{DA}$
$\therefore\text{SP }||\text{ RQ}$ and $\therefore\text{PQ }||\text{ SR}$ and $\text{PQ}=\text{RQ}=\text{SP}=\text{SR}$
Hence, PQRS is a rhombus.
View full question & answer→
Question 403 Marks
In the given figure, O is a point inside a $\triangle\text{PQR}$ such that $\angle\text{PQR}=90^\circ,\text{OP}=6\text{cm}$ and $\text{OR}=8\text{cm}.$ If $\text{PQ}=24\text{cm}$ and $\text{QR}=26\text{cm},$ prove that $\triangle\text{PQR}$ is right-angled.
Answer
In $\triangle\text{PQR},\angle\text{QPR}=90^\circ,\text{PQ}=24\text{cm},$ and $\text{QR}=26\text{cm}^2$ In $\triangle\text{POR},\text{PO}=6\text{cm},\text{QR}=8\text{cm},$ and $\angle\text{POR}=90^\circ$
In $\triangle\text{POR},$ $\text{PR}^2=\text{PO}^2+\text{OR}^2$ $\text{PR}^2=(6^2+8^2)\text{cm}^2$ $=(36+64)\text{cm}^2=100\text{cm}^2$ $\text{PR}=\sqrt{100}\text{cm}=10\text{cm}$ In $\triangle\text{PQR},$ By Pythagoras theoram, we have $\text{QR}^2=\text{QP}^2+\text{PR}^2$ $(26)^2\text{cm}^2=\Big(24^2+10^2\Big)\text{cm}^2$ $676\text{cm}^2=(576+100)\text{cm}^2$ $676\text{cm}^2=676\text{cm}^2$ Hence, $\text{QR}^2=\text{QP}^2+\text{PR}^2$(sum of square of two sides equal to square of greatest side) Hence, $\triangle\text{PQR}$ is a right triangle which ois right angled at P.
View full question & answer→
Question 413 Marks
In the given pairs of triangles, find which pair of triangles are similar. State the similarity criterior and write the similarity relation in symbolic from.
Answer
In $\triangle\text{EFD}$ and $\triangle\text{PQR}$
FE = 2cm, FD = 3cm, ED = 2.5cm
PQ = 4cm, PR = 6cm, QR = 5cm
$\Rightarrow\frac{\text{PQ}}{\text{FE}}=\frac{\text{PR}}{\text{FD}}=\frac{\text{QR}}{\text{ED}}$
$\therefore\triangle\text{FED}\sim\triangle\text{PQR}$ (SSS similarity)
View full question & answer→
Question 423 Marks
The corresponding sides of two similar triangles are in the ratio $2 : 3.$ If the area of the smaller triangle is $48\ cm^2,$ find the area of the larger triangle.
Answer
Given that the triangles are similar.
$\Rightarrow\frac{\text{area of the smaller triangle}}{\text{area of the larger triangle}}=\Big(\frac{2}{3}\Big)^2$
$\Rightarrow\frac{48}{\text{area of the larger triangle}}=\frac{4}{9}$
$\Rightarrow\text{area of the larger triangle = 108cm}^2$
View full question & answer→
Question 433 Marks
A vertical pole of lenght 7.5m casts a shadow 5m long on the ground and at the same time a tower casts a shadow 24m long. Find the height of the tower.
Answer

Let AB be the vertical stick and let AC be its shadow.
Then, AB = 7.5m and AC = 5m
Let DE be the vertical tower and let DF be its shadow
Then, DF = 24m, Let DE = x meters
Now, In $\triangle\text{BAC}$ and $\triangle\text{EDF},$
$\triangle\text{BAC}\sim\triangle\text{EDF}$ by SAS criterion
$\Rightarrow\frac{\text{AB}}{\text{DE}}=\frac{\text{AC}}{\text{DF}}$
$\Rightarrow\frac{7.5}{\text{x}}=\frac{5}{24}$
$\Rightarrow\text{x}=\frac{7.5\times24}{5}=36\text{m}$
therefore, height of the vertical tower is 36m.
View full question & answer→
Question 443 Marks
In a right triangle $ABC,$ right-angled at $B, D$ is a point on hypotenuse such that $\text{BD}\perp\text{AC}.$ If $\text{DP}\perp\text{AB}$ and $\text{DQ}\perp\text{BC}$ then prove that.
  1. $\text{DQ}^2=\text{DP}.\text{DQ}$
  2. $\text{DP}^2=\text{DQ}.\text{AP}$
Answer
We know that
when a perpendicular is drawn from the vertex of a triangle on to the hypotenuse, then the triangles on both sides of the perpendicular are similar to eachother and the to the whole triangle.
  1. In $\triangle\text{DBC},$
$\triangle\text{DQB}\sim\triangle\text{DQC}$
$\Rightarrow\frac{\text{DQ}}{\text{DQ}}=\frac{\text{QB}}{\text{QC}}$
$\Rightarrow\text{DQ}^2=\text{QB}.\text{QC}$
Since all the angles of PBQD are $90^\circ ,$
$PBQD$ is a rectangle.
$\Rightarrow QB = DP$ and $PB = DQ ....(i)$
$\Rightarrow DQ^2 = DP . QC$
  1. Similarly, since $PD$ is a perpendicular on $AB$,
$\triangle\text{APD}\sim\triangle\text{DPB}$
$\Rightarrow\frac{\text{DP}}{\text{PB}}=\frac{\text{AP}}{\text{DP}}$
$\Rightarrow\frac{\text{DP}}{\text{DQ}}=\frac{\text{AP}}{\text{DP}}\dots(\text{using }(\text{i}))$
$\Rightarrow\text{DP}^2=\text{DQ}.\text{AP}$
 
View full question & answer→
Question 453 Marks
In $\triangle\text{ABC},$ the bisector of $\angle\text{B}$ meets AC at D. A line PQ || AC meets AB, BC and BD at P, Q and R respectively.
Show that PR × BQ = QR × BP.
Answer
In triangle BQP, BR bisects angle B.
Applying angle bisector theorem, we get:
$\frac{\text{QR}}{\text{PR}}=\frac{\text{BQ}}{\text{BP}}$
⇒ BP × QR = BQ × PR
This completes the proof.
View full question & answer→
Question 463 Marks
In the given pairs of triangles, find which pair of triangles are similar. State the similarity criterior and write the similarity relation in symbolic from.
Answer
In $\triangle\text{ABC}$ and $\triangle\text{EFD}$
$\angle\text{A}=\angle\text{D}=70^\circ$
SAS: Similarity condition is not satisfied as $\angle\text{A}$ and $\angle\text{D}$ are not included angles.
View full question & answer→
Question 473 Marks
In the given figure, $\angle\text{CAB}=90^\circ$ and $\text{AD}\perp\text{BC}.$ Show that $\triangle\text{BDA}\sim\triangle\text{BAC}.$ If AC = 75cm, AB = 1m, and BC = 1,25m find AD.
Answer

Given: AB = 100cm, BC = 125cm, AC = 75cm
Proof:
In $\triangle\text{BAC}$ and $\triangle\text{BDA}$
$\angle\text{BAC}=\angle\text{BDA}=90^\circ$
$\angle\text{B}=\angle\text{B}$ (common)
$\triangle\text{BAC}\sim\triangle\text{BDA}$ (by AA similarities)
$\Rightarrow\frac{\text{BA}}{\text{BC}}=\frac{\text{AD}}{\text{AC}}$
$\Rightarrow\frac{\text{100}}{\text{125}}=\frac{\text{AD}}{\text{75}}$
$\Rightarrow\text{AD}=\frac{\text{100}\times75}{\text{125}}=60\text{ cm}$
Therefore, AD = 60cm
View full question & answer→
Question 483 Marks
In the given figure, DE || BC such that AD = x cm, DB = (3x + 4)cm, AE = (x + 3)cm and EC = (3x + 19)cm. Find the value of x.
Answer
In $\triangle\text{ABC,}$
DE || BC
By Basic proportionality theorem,
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{\text{x}}{\text{3x}+4}=\frac{\text{x}+3}{\text{3x}+19}$
$\Rightarrow\text{x}(3\text{x}+19)=(\text{x}+3)(3\text{x}+4)$
$\Rightarrow3\text{x}^2+19\text{x}=3\text{x}^2+4\text{x}+9\text{x}+12$
$\Rightarrow6\text{x}=12$
$\Rightarrow\text{x}=2$
View full question & answer→
Question 493 Marks
Find the lenght of each side of a rhombus whose diagonals are 24cm and 10cm. long.
Answer
Let ABCD be the given rhombus whose diagonals intersect at O. Then AC = 24cm and BD = 10cm
We know that the diagonals of a rhombus bisect each other at right angles. $\text{OA}=\frac{1}{2}\text{AC}=12\text{cm}$ $\text{OB}=\frac{1}{2}\text{BD}=5\text{cm}$ and $\angle\text{AOB}=90^\circ$ Form right $\triangle\text{AOB},$ we have $\text{AB}^2=\text{OA}^2+\text{OB}^2$ $\Rightarrow\text{AB}^2=\Big[(12)^2+(5)^2\Big]\text{cm}^2$ $\Rightarrow(144+25)\text{cm}^2=169\text{cm}^2$ $\Rightarrow\text{AB}=\sqrt{169}\text{cm}=13\text{cm}$ Hence, each side of a rhombus 13cm.
View full question & answer→
Question 503 Marks
In a $\triangle\text{ABC},\text{AD}$ is the bisector of $\angle\text{A}.$
If AB = 5.6cm, AC = 4cm and DC = 3cm, find BC.
Answer
It is given that AD bisects $\angle\text{A}.$
Applying angle-bisector theorem in $\triangle\text{ABC},$ we get:
$\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\frac{\text{BD}}{3}=\frac{5.6}{\text{4}}$
$\Rightarrow\text{BD}=\frac{5.6\times3}{4}$
$\Rightarrow\text{BD}=4.2\text{cm}$
Hence, BC = 3 + 4.2 = 7.2cm
View full question & answer→
Question 513 Marks
Find the length of altitude AD of an isosceles $\triangle\text{ABC}$ in which AB = AC = 2a units and BC = a units.
Answer
Given: $\triangle\text{ABC}$ in which AB = AC = 2a units and BC = a units. Const: Draw $\text{AD}\perp\text{BC}$ then D is the midpoint of BC.
In $\triangle\text{ABC}$ $\text{BC}=\text{a}$ and $\text{BD}=\frac{\text{BC}}{2}=\frac{\text{a}}{2}$ In $\triangle\text{ADB},$ $(\text{AB})^2=\text{AD}^2+\text{BD}^2$ $\text{AD}^2=\Big(\text{AB}^2-\text{BD}^2\Big)$ $\text{AD}^2=\bigg[(\text{2a})^2-\Big(\frac{\text{a}}{2}\Big)^2\bigg]$ $\text{AD}^2=\bigg[4\text{a}^2-\frac{\text{a}^2}{4}\bigg]=\frac{15\text{a}^2}{4}$ $\Rightarrow\text{AD}=\frac{\text{a}\sqrt{15}}{2}\text{units}$
View full question & answer→
Question 523 Marks
If the lengths of the sides BC, CA and AB of a $\triangle\text{ABC}$ are a, b and c respectively and AD is the bisectore of $\angle\text{A}$ then find the lengths of BD and DC.
Answer

Given $\triangle\text{ABC}$ in which AD, the bisector of $\angle\text{A}$ meets BC in D.
Let = x ⇒ DC = (a - x)
Then by the angle Bisector theorem,
$\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\frac{\text{x}}{\text{a}-\text{x}}=\frac{\text{c}}{\text{b}}$
$\Rightarrow\text{xb}=\text{c}(\text{a}-\text{x})$
$\Rightarrow\text{xb}=\text{ac}-\text{xc}$
$\Rightarrow\text{xb}+\text{xc}=\text{ac}$
$\Rightarrow\text{x}=\frac{\text{ac}}{\text{b+c}}$
So, $\text{BD}=\frac{\text{ac}}{\text{b+c}}$
$\text{DC}=\text{a}-\text{x}\Rightarrow\text{DC}=\text{a}-\frac{\text{ac}}{\text{b+c}}$
$=\frac{\text{ab}+\text{ac}-\text{ac}}{\text{b}+\text{c}}=\frac{\text{ab}}{\text{b}+\text{c}}$
Hence, $\text{BD}=\frac{\text{ac}}{\text{b}+\text{c}}$ and $\text{DC}=\frac{\text{ab}}{\text{b}+\text{c}}.$
View full question & answer→
Generate Paper FreeAll Triangles topics