Numerical: geometry MCQ - Practice Questions with Answers

The figure marks angle PQR as 50°. Since QR = PR, triangle PQR is isosceles, so the angles opposite these equal sides are equal; hence angle QPR is also 50°. The remaining angle QRP is 180° - 50° - 50° = 80°. Because QR is parallel to PM, angle RPM equals the corresponding angle between PR and QR, which is 80°. The values 70°, 100° and 95° do not satisfy both the isosceles-triangle condition and the parallel-line angle relation.

Let AB = BC = x. Then AC = x√2, so the perimeter is x(2 + √2). The given perimeter is 14 + 7√2 = 7(2 + √2), hence x = 7 cm. The sides are 7, 7, and 7√2; since (7√2)² = 7² + 7², the triangle is right-angled at B. Its area is 1/2 × 7 × 7 = 49/2 cm². The values 59 cm², 69 cm², and 75/2 cm² do not follow from the side lengths fixed by the perimeter.

Let the base be b cm. Each equal side is \frac{5b}{6}, so the perimeter is b+2\times\frac{5b}{6}=\frac{8b}{3}=32, giving b=12 cm and each equal side 10 cm. The height bisects the base, so it is \sqrt{10^2-6^2}=8 cm. The area is \frac{1}{2}\times12\times8=48 cm2, and half of this area is 24 cm2. Thus 38, 42, and 48 are not the required half-area.

Let the breadth of the rectangle be x metres, so its length is 2x metres. The perimeter is 2(length + breadth), so 2(2x + x) = 90. This gives 6x = 90 and x = 15. Therefore, the breadth is 15 m and the length is 30 m. The area is length × breadth = 30 × 15 = 450 m2. The values 380 m2, 460 m2 and 420 m2 do not satisfy the rectangle dimensions obtained from the given perimeter and ratio.