Question 16
In the given figure, PQ $⊥$ RQ, PQ = 5 cm and QR = 5 cm. Then, $△ P Q R$ is

a) a right-angled triangle but not isosceles
b) an isosceles right-angled triangle
c) isosceles but not a right-angled triangle
d) neither isosceles nor right-angled triangle

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Solution

Given : PQ $⊥$ RQ, PQ = 5 cm and QR = 5 cm

Since, PQ $⊥$ RQ

So, $∠$ PQR = $90^{\circ}$

$∴ △$ PQR is a right angled triangle.

Also, in $△$ PQR,

PQ = QR

$∴ △$ PQR is an isosceles triangle.

Hence, $△$ PQR is an isosceles right-angled triangle.

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Similar questions

Question 16
In the given figure, PQ $⊥$ RQ, PQ = 5 cm and QR = 5 cm. Then, $△ P Q R$ is

a) a right-angled triangle but not isosceles
b) an isosceles right-angled triangle
c) isosceles but not a right-angled triangle
d) neither isosceles nor right-angled triangle

Question 7

Lengths of sides of a triangle are 3 cm, 4 cm and 5 cm. The triangle is

a) Obtuse-angled triangle
b) Acute angled triangle
c) Right angled triangle
d) An isosceles right triangle.

Q. In ∆ABC and ∆PQR, it is given that AB = AC, ∠C = ∠P and ∠B = ∠Q. Then, the two triangles are
(a) isosceles but not congruent
(b) isosceles and congruent
(c) congruent but not isosceles
(d) neither congruent nor isosceles

Q. PQR is a right angled isosceles triangle, right angle at Q, then prove that

P R^{2} = 2 P Q^{2}

Q. If

△ P Q R

is an isosceles right triangle, right-angled at

R

. Prove that

{P Q}^{2} = 2 {P R}^{2}

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Isosceles Triangle

Standard VII Mathematics

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Question 16 In the given figure, PQ⊥RQ, PQ = 5 cm and QR = 5 cm. Then, △PQR is a) a right-angled triangle but not isosceles b) an isosceles right-angled triangle c) isosceles but not a right-angled triangle d) neither isosceles nor right-angled triangle

Given : PQ⊥RQ, PQ = 5 cm and QR = 5 cm Since, PQ⊥RQ So, ∠ PQR = 90∘ ∴△ PQR is a right angled triangle. Also, in △PQR, PQ = QR ∴△ PQR is an isosceles triangle. Hence, △ PQR is an isosceles right-angled triangle.

Question 16
In the given figure, PQ $⊥$ RQ, PQ = 5 cm and QR = 5 cm. Then, $△ P Q R$ is

a) a right-angled triangle but not isosceles
b) an isosceles right-angled triangle
c) isosceles but not a right-angled triangle
d) neither isosceles nor right-angled triangle

Given : PQ $⊥$ RQ, PQ = 5 cm and QR = 5 cm

Since, PQ $⊥$ RQ

So, $∠$ PQR = $90^{\circ}$

$∴ △$ PQR is a right angled triangle.

Also, in $△$ PQR,

PQ = QR

$∴ △$ PQR is an isosceles triangle.

Hence, $△$ PQR is an isosceles right-angled triangle.