IPMPrepFind the circumradius of a right triangle whose legs are 14 cm and 48 cm.
Solution
Hypotenuse = 50. The hypotenuse is a diameter of the circumcircle.
R = 502 = 25 cm.
Answer: (C) 25 cm
Quantitative Ability · Geometry
Geometry: IPMAT practice questions
24 questions, 6 of them short answer (type the number, as in IPMAT Indore). Each is marked with the exam level it suits. Answer, then open the step-by-step solution.
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A convex polygon has 135 diagonals. Find the sum of its interior angles in degrees.
Solution
n(n − 3)/2 = 135 gives n = 18.
Sum of interior angles = (n − 2) × 180° = 2880°.
Answer: 2880
In triangle ABC, AB = 6 cm, AC = 9 cm and BC = 12 cm. The bisector of angle A meets BC at D. Find BD.
Solution
The angle bisector divides the opposite side in the ratio of the other two sides: BD : DC = AB : AC.
BD = 12 × 6/(6 + 9) = 4.8 cm.
Answer: (A) 4.8 cm
A chord of a circle of radius 5 cm is 3 cm from the centre. Find the length of the chord.
Solution
The perpendicular from the centre bisects the chord: half-chord = √(52 − 32) = 4.
Chord = 2 × 4 = 8 cm.
Answer: (A) 8 cm
In triangle ABC, DE is parallel to BC with D on AB and AD : DB = 3 : 4. If the area of triangle ABC is 196 sq cm, find the area of trapezium DECB.
Solution
Triangle ADE is similar to ABC with ratio AD : AB = 3 : 7.
Area ADE = 196 × (37)2 = 36.
Area DECB = 196 − 36 = 160 sq cm.
Answer: (B) 160 sq cm
Find the inradius of a triangle with sides 5 cm, 5 cm and 6 cm.
Solution
s = 8; area by Heron's formula = 12 sq cm.
r = area/s = 128 = 1.5 cm.
Answer: (A) 1.5 cm
Find the inradius of a triangle with sides 20 cm, 34 cm and 42 cm.
Solution
s = 48; area by Heron's formula = 336 sq cm.
r = area/s = 33648 = 7 cm.
Answer: (D) 7 cm
Find the inradius of a right triangle whose legs are 7 cm and 24 cm.
Solution
Hypotenuse = √(72 + 242) = 25.
For a right triangle, r = (a + b − c)/2 = (7 + 24 − 25)/2 = 3 cm.
Answer: (D) 3 cm
ABCD is a cyclic quadrilateral with ∠A = 60°. Find ∠C.
Solution
Opposite angles of a cyclic quadrilateral add to 180°.
∠C = 180° − 60° = 120°.
Answer: (A) 120°
In triangle ABC, AB = 4 cm, AC = 8 cm and BC = 8 cm. The bisector of angle A meets BC at D. Find BD.
Solution
The angle bisector divides the opposite side in the ratio of the other two sides: BD : DC = AB : AC.
BD = 8 × 4/(4 + 8) = 2 23 cm.
Answer: (D) 2 23 cm
In triangle ABC, DE is parallel to BC with D on AB and AD : DB = 1 : 4. If the area of triangle ABC is 120 sq cm, find the area of trapezium DECB.
Solution
Triangle ADE is similar to ABC with ratio AD : AB = 1 : 5.
Area ADE = 120 × (15)2 = 4.8.
Area DECB = 120 − 4.8 = 115.2 sq cm.
Answer: (B) 115.2 sq cm
In triangle ABC, AB = 5 cm, AC = 15 cm and BC = 14 cm. The bisector of angle A meets BC at D. Find BD.
Solution
The angle bisector divides the opposite side in the ratio of the other two sides: BD : DC = AB : AC.
BD = 14 × 5/(5 + 15) = 3.5 cm.
Answer: (A) 3.5 cm
How many non-congruent triangles with integer sides have a perimeter of 15?
Solution
List sides a ≤ b ≤ c with a + b + c = 15 and a + b > c (so c < 152).
They are: (1, 7, 7), (2, 6, 7), (3, 5, 7), (3, 6, 6), (4, 4, 7), (4, 5, 6), (5, 5, 5).
Count = 7.
Answer: 7
Find the area of a triangle with sides 10 cm, 10 cm and 16 cm.
Solution
s = (10 + 10 + 16)/2 = 18.
Area = √(18 × 8 × 8 × 2) = 48 sq cm.
Answer: 48
Find the inradius of a right triangle whose legs are 18 cm and 24 cm.
Solution
Hypotenuse = √(182 + 242) = 30.
For a right triangle, r = (a + b − c)/2 = (18 + 24 − 30)/2 = 6 cm.
Answer: (D) 6 cm
How many diagonals does a convex polygon with 6 sides have?
Solution
Each vertex joins to n − 3 others by a diagonal, and each diagonal is counted twice.
n(n − 3)/2 = 6 × 32 = 9.
Answer: 9
How many non-congruent isosceles triangles (including equilateral) with integer sides have a perimeter of 14?
Solution
List sides a ≤ b ≤ c with a + b + c = 14 and a + b > c (so c < 7).
They are: (2, 6, 6), (4, 4, 6), (4, 5, 5).
Count = 3.
Answer: (C) 3
How many non-congruent isosceles triangles (including equilateral) with integer sides have a perimeter of 12?
Solution
List sides a ≤ b ≤ c with a + b + c = 12 and a + b > c (so c < 6).
They are: (2, 5, 5), (4, 4, 4).
Count = 2.
Answer: 2
ABCD is a cyclic quadrilateral with ∠A = 83°. Find ∠C.
Solution
Opposite angles of a cyclic quadrilateral add to 180°.
∠C = 180° − 83° = 97°.
Answer: (C) 97°
AB is a diameter of a circle and C is a point on the circle with ∠CAB = 65°. Find ∠CBA.
Solution
The angle in a semicircle is 90°, so ∠ACB = 90°.
∠CBA = 180° − 90° − 65° = 25°.
Answer: (B) 25°
In triangle ABC, DE is parallel to BC with D on AB and AD : DB = 1 : 1. If the area of triangle ABC is 72 sq cm, find the area of triangle ADE.
Solution
AD : AB = 1 : 2; areas of similar triangles are in the ratio of the squares of their sides.
Area ADE = 72 × 14 = 18 sq cm.
Answer: (D) 18 sq cm
A chord of a circle of radius 20 cm is 12 cm from the centre. Find the length of the chord.
Solution
The perpendicular from the centre bisects the chord: half-chord = √(202 − 122) = 16.
Chord = 2 × 16 = 32 cm.
Answer: (A) 32 cm
How many diagonals does a convex polygon with 9 sides have?
Solution
Each vertex joins to n − 3 others by a diagonal, and each diagonal is counted twice.
n(n − 3)/2 = 9 × 62 = 27.
Answer: 27
From a point 13 cm from the centre of a circle of radius 5 cm, a tangent is drawn. Find the length of the tangent.
Solution
The radius to the point of contact is perpendicular to the tangent.
Tangent = √(132 − 52) = 12 cm.
Answer: (A) 12 cm
These are original practice questions written by Target Campus in the pattern, syllabus and difficulty of IPMAT Indore, IPMAT Rohtak, JIPMAT, NPAT and SET. They are not previous-year papers.