In ∆TPQ, if ∠T = 65°, ∠P = 95° , Which of the following is a true statement?
- APQ < TP
- ✓PQ < TQ
- CTQ < TP < PQ
- DPQ < TP < TQ
Answer: B.
View full solution →Question types
Triangles question types
68 questions across 7 question groups — pick any mix to generate a Maths paper with step-by-step answer keys.
68
Questions
7
Question groups
5
Question types
01
MCQ(1M)
3 Q→021 Mark Question
6 Q→032 Mark Question
26 Q→043 Mark Question
18 Q→054 Mark Question
6 Q→065 Mark Question
7 Q→07Activity Based Questions.
2 Q→Sample Questions
Triangles questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
In ∆PQR, if ∠R > ∠Q, then _____ .
- AQR > PR
- ✓PQ > PR
- C3.8 cm
- D3.4 cm
Answer: B.
View full solution →If two sides of a triangle are 5 cm and 1.5 cm, the length of its third side cannot be ____.
- A3.7 cm
- B4.1 cm
- C3.8 cm
- ✓3.4 cm
Answer: D.
View full solution →On a card-sheet, draw a triangle of sides 4 cm, 3 cm and 2 cm. Cut it out. Make 13 more copies of the triangle and cut them out from the card sheet. Note that all these triangular pieces are congruent. Arrange them as shown in the following figure and make three triangles out of them.
View full solution →
Number of triangle: 1
Number of triangles: 4
Number of triangles: 9
∆ABC and ∆DEF are similar in the correspondence ABC ↔ DEF.
∠A ≅ ∠D, ∠B ≅ ∠E, ∠C ≅ ∠F
and $\frac{ AB }{ DE }=\frac{4}{8}=\frac{1}{2} ; \frac{ BC }{ EF }=\frac{3}{6}=\frac{1}{2} ; \frac{ AC }{ DF }=\frac{2}{4}=\frac{1}{2} \ldots$ the corresponding sides are inproportion.
Similarly, consider ∆DEF and ∆PQR. Are their angles congruent and sides proportional in the correspondence DEF ↔ PQR?
Draw a triangle ABC on a cardboard. Draw its medians and denote their point of concurrence as G. Cut out the triangle. Now take a pencil. Try to balance the triangle on the flat tip of the pencil. The triangle is balanced only when the point G is on the flat tip of the pencil. This activity shows an important property of the centroid (point of concurrence of the medians) of the triangle. Point it out.
Draw a triangle ABC. Draw medians AD, BE and CF of the triangle. Let their point of concurrence be G, which is called the centroid of the triangle. Compare the lengths of AG and GD with a divider. Verify that the length of AG is twice the length of GD. Similarly, verify that the length of BG is twice the length of GE and the length of CG is twice the length of GF. Name the property of medians of a triangle observed here.
Every student in the group should draw a right angled triangle, one of the angles measuring 30°. The choice of lengths of sides should be their own. Each one should measure the length of the hypotenuse and the length of the side opposite to 30° angle.
One of the students in the group should fill in the following table.
One of the students in the group should fill in the following table.
Did you notice any property of sides of right angled triangle with one of the angles measuring 30°?
The measures of angles of a triangle are in the ratio 5 : 6 : 7. Find the measures.
Draw a triangle $\Delta A _1 B_1 C _1$ on a card-sheet and cut it out.
Measure $\angle A _1, \angle B_1, \angle C _1$
Draw two more triangles $AA _2 B_2 C _2$ and $AA _3 B_3 C _3$ such that
$\angle A _1=\angle A _2=\angle A _3, \angle B_1=\angle B _2=\angle B _3, \angle C _1=\angle C _2=\angle C _3$ and $B _1 C _1> B _2 C _2> B _3 C _3$.
Now cut these two triangles also.
Measure the lengths of the three triangles. Arrange the triangles in two ways as shown in the figure.
Check the ratios $\frac{A_1 B_1}{A_2 B_2}, \frac{B_1 C_1}{B_2 C_2}, \frac{A_1 C_1}{A_2 C_2}$.
You will notice that the ratios are equal.
Similarly, see whether the ratios $\frac{A_1 C_1}{A_3 C_3}, \frac{B_1 C_1}{B_3 C_3}, \frac{A_1 B_1}{A_3 B_3}$ are equal.
What do you observe?
View full solution →Measure $\angle A _1, \angle B_1, \angle C _1$
Draw two more triangles $AA _2 B_2 C _2$ and $AA _3 B_3 C _3$ such that
$\angle A _1=\angle A _2=\angle A _3, \angle B_1=\angle B _2=\angle B _3, \angle C _1=\angle C _2=\angle C _3$ and $B _1 C _1> B _2 C _2> B _3 C _3$.
Now cut these two triangles also.
Measure the lengths of the three triangles. Arrange the triangles in two ways as shown in the figure.
Check the ratios $\frac{A_1 B_1}{A_2 B_2}, \frac{B_1 C_1}{B_2 C_2}, \frac{A_1 C_1}{A_2 C_2}$.
You will notice that the ratios are equal.
Similarly, see whether the ratios $\frac{A_1 C_1}{A_3 C_3}, \frac{B_1 C_1}{B_3 C_3}, \frac{A_1 B_1}{A_3 B_3}$ are equal.
What do you observe?
The measures of angles of a set square in your compass box are 30°, 60° and 90°. Verify the property of the sides of the set square.
Draw a triangle of any measure on a thick paper. Take a point T on ray QR as shown in the figure given below. Cut two pieces of thick paper which will exactly fit the comers of ∠P and ∠Q. See that the same two jpieces fit exactly at the comer of ∠PRT as shown in the figure.
In the adjoining figure, seg PT is the bisector of ∠QPR. A line through R intersects ray QP at point S. Prove that PS = PR.
Given: Seg PT is the bisector of ∠QPR.
To prove: PS = PR
Construction: Draw seg SR || seg PT.
In the adjoining figure, bisector of ∠B AC intersects side BC at point D. Prove that AB > BD.
Given: Bisector of ∠BAC intersects side BC at point D.
To prove: AB > BD
In the adjoining figure, points D and E are on side BC of ∆ABC, such that BD = CE and AD AE. Show that ∆ABD ≅ ∆ACE.
Given: Points D and E are on side BC of ∆ABC,
such that BD = CE and AD = AE.
To prove: ∆ABD ≅ ∆ACE
In ∆PQR, if PQ > PR and bisectors of ∠Q and ∠R intersect at S. Show that SQ > SR.
Given: In APQR, PQ > PR and bisectors of ∠Q and ∠R intersect at S.
To prove: SQ > SR
Given: In APQR, PQ > PR and bisectors of ∠Q and ∠R intersect at S.
To prove: SQ > SR
Some information is shown in △ ABC and △ PQR in figure 3.57. Observe it. Hence find the lengths of side AC and PQ.
Prove the Theorem : The sum of any two sides of a triangle is greater than the third side.
In figure 3.7, bisectors of ∠B and ∠C of D ABC intersect at point P.
Prove that ∠BPC = 90 + $\frac{1}{2}$ ∠BAC.
View full solution →Prove that ∠BPC = 90 + $\frac{1}{2}$ ∠BAC.
In the adjoining figure, point S is any point on side QR of ∆PQR. Prove that: PQ + QR + RP > 2PS
We have learnt that if two triangles are equiangular then their sides are in proportion. What do you think if two quadrilaterals are equiangular? Are their sides in proportion? Draw different figures and verify. Verify the same for other polygons.
Find the values of x and y using the information shown in the given figure. Find the measures of ∠ABD and ∠ACD.
In the given figure, bisectors of $\angle B$ and $\angle C$ of $\triangle A B C$ intersect at point P. Prove that $\angle BPC =90^{\circ}+\frac{1}{2} \angle B A C$.
Complete the proof by filling in the blanks.
View full solution →Complete the proof by filling in the blanks.
In the adjoining figure, measures of some angles are given. Using the measures, find the values of x, y, z.
Check the congruence of triangles.
In the adjoining figure, seg $A D \perp$ seg $B C$. Seg $A E$ is the bisector of $\angle C A B$ and $B-E-C$. Prove that $\angle DAE =\frac{1}{2}(\angle C -\angle B )$.
Given: seg AD ⊥ seg BC
seg AE is the bisector of ∠CAB.
To prove: $\angle DAE =\frac{1}{2}(\angle C-\angle B)[\because AD \perp BC ]$
∴ ∠DAE = 180° – 90° – ∠AED
∴ ∠DAE = 90° – ∠AED ….(ii)
View full solution →Given: seg AD ⊥ seg BC
seg AE is the bisector of ∠CAB.
To prove: $\angle DAE =\frac{1}{2}(\angle C-\angle B)[\because AD \perp BC ]$
∴ ∠DAE = 180° – 90° – ∠AED
∴ ∠DAE = 90° – ∠AED ….(ii)
∆ABC is isosceles in which AB = AC. Seg BD and seg CE are medians. Show that BD = CE.
Given: In isosceles ∆ABC, AB = AC. seg BD and seg CE are the medians of ∆ABC.
To prove: BD = CE
Observe the information shown in pairs of triangles given below. State the test by which the two triangles are congruent. Write the remaining congruent parts of the triangles.
In each of the examples given below, a pair of triangles is shown. Equal parts of triangles in each pair are marked with the same signs. Observe the figures and state the test by which the triangles in each pair are congruent.
Prove the Theorem : If two angles of a triangle are congruent then the sides opposite to them are congruent.
Prove the Theorem : If two sides of a triangle are congruent then the angles opposite to them are congruent.
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