# CAT Geometry Triangles Questions PDF [Most Important]

**Geometry** **Triangles** questions are important concepts in the Geometry concept of the **CAT** **Quant **section. These questions are not very tough; make sure you are aware of all the **Important Formulas in CAT Geometry**. Solve more questions from CAT **Geometry**** Triangles**. You can check out these CAT Geometry questions from the **CAT Previous year papers**. Practice a good number of questions in CAT Geometry **Triangles** so that you can answer these questions with ease in the exam. In this post, we will look into some important CAT Geometry Questions. These are a good source of practice for CAT 2022 preparation; If you want to practice these questions, you can download these Important** Triangles (Geometry) Questions for CAT** (with detailed answers) **PDF** along with the video solutions below, which is completely Free.

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**Question 1: **In a triangle ABC, the lengths of the sides AB and AC equal 17.5 cm and 9 cm respectively. Let D be a point on the line segment BC such that AD is perpendicular to BC. If AD = 3 cm, then what is the radius (in cm) of the circle circumscribing the triangle ABC?

a) 17.05

b) 27.85

c) 22.45

d) 32.25

e) 26.25

**1) Answer (E)**

**Solution:**

Let x be the value of third side of the triangle. Now we know that Area = 17.5*9*x/(4*R), where R is circumradius.

Also Area = 0.5*x*3 .

Equating both, we have 3 = 17.5*9 / (2*R)

=> R = 26.25.

**Question 2: **Consider obtuse-angled triangles with sides 8 cm, 15 cm and x cm. If x is an integer then how many such triangles exist?

a) 5

b) 21

c) 10

d) 15

e) 14

**2) Answer (C)**

**Solution:**

For obtuse-angles triangle, $c^2 > a^2 + b^2$ and c < a+b

If 15 is the greatest side, 8+x > 15 => x > 7 and $225 > 64 + x^2$ => $x^2$ < 161 => x <= 12

So, x = 8, 9, 10, 11, 12

If x is the greatest side, then 8 + 15 > x => x < 23

$x^2 > 225 + 64 = 289$ => x > 17

So, x = 18, 19, 20, 21, 22

So, the number of possibilities is 10

**Question 3: **If the lengths of diagonals DF, AG and CE of the cube shown in the adjoining figure are equal to the three sides of a triangle, then the radius of the circle circumscribing that triangle will be?

a) equal to the side of the cube

b) $\sqrt 3$ times the side of the cube

c) 1/$\sqrt 3$ times the side of the cube

d) impossible to find from the given information

**3) Answer (A)**

**Solution:**

Consider side of the cube as x.

So diagonal will be of length $\sqrt{3}$ * x.

Now if diagonals are side of equilateral triangle we get area = 3*$\sqrt{3}*x^2$ /4 .

Also in a triangle

4 * Area * R = Product of sides

4* 3*$\sqrt{3}*x^2$ /4 * R = .3*$\sqrt{3}*x^3$

R = x

**Question 4: **In triangle DEF shown below, points A, B and C are taken on DE, DF and EF respectively such that EC = AC and CF = BC. If angle D equals 40 degress , then angle ACB is ?

a) 140

b) 70

c) 100

d) None of these

**4) Answer (C)**

**Solution:**

Let angle EAC = x, so angle AEC = x and angle ACE = 180-2x

Let angle FBC = y, so angle BFC = y and angle BCF = 180-2y

So, angle ACB = 180-(180-2x+180-2y) = 2(x+y) – 180

x+y = 180 – 40 = 140

So, angle ACB = 280 – 180 = 100

**Question 5: **If a,b,c are the sides of a triangle, and $a^2 + b^2 +c^2 = bc + ca + ab$, then the triangle is:

a) equilateral

b) isosceles

c) right angled

d) obtuse angled

**5) Answer (A)**

**Solution:**

$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) => 3(a^2 + b^2 + c^2)$

This is possible only if a = b = c.

So, the triangle is an equilateral triangle.

**Question 6: **In the figure given below, ABCD is a rectangle. The area of the isosceles right triangle ABE = 7 $cm^2$ ; EC = 3(BE). The area of ABCD (in $cm^2$) is

a) 21 $cm^2$

b) 28 $cm^2$

c) 42 $cm^2$

d) 56 $cm^2$

**6) Answer (D)**

**Solution:**

Let AB = BE = x

Area of triangle ABE = $x^2/2$ = 14; we get x = $\sqrt{14}$

So we have side BC = 4*$\sqrt{14}$

Now area is AB*BC = 14 *4 = 56 $cm^2$

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**Question 7: **In the figure given below, ABCD is a rectangle. The area of the isosceles right triangle ABE = 7 $cm^2$ ; EC = 3(BE). The area of ABCD (in $cm^2$) is

a) 21 $cm^2$

b) 28 $cm^2$

c) 42 $cm^2$

d) 56 $cm^2$

**7) Answer (D)**

**Solution:**

Let AB = BE = x

Area of triangle ABE = $x^2/2$ = 14; we get x = $\sqrt{14}$

So we have side BC = 4*$\sqrt{14}$

Now area is AB*BC = 14 *4 = 56 $cm^2$

**Question 8: **The area of the triangle whose vertices are (a,a), (a + 1, a + 1) and (a + 2, a) is

[CAT 2002]

a) $a^3$

b) $1$

c) $2a$

d) $2^{1/2}$

**8) Answer (B)**

**Solution:**

The triangle we have is :

The length of three sides is $\sqrt 2, \sqrt 2$ and $2$.

This is a right-angled triangle.

Hence, it’s area equals $1/2 * \sqrt 2 * \sqrt 2 = 1$

So, the correct answer is b)

Alternate Approach :

Area of triangle = $\frac{1}{2}\times\ base\times\ height$

So we get $\frac{1}{2}\times2\times\ 1$=1 square units

**Question 9: **In the following figure, ACB is a right-angled triangle. AD is the altitude. Circles are inscribed within the triangle ACD and triangle BCD. P and Q are the centers of the circles. The distance PQ is

The length of AB is 15 m and AC is 20 m

a) 7 m

b) 4.5 m

c) 10.5 m

d) 6 m

**9) Answer (A)**

**Solution:**

By Pythagoras theorem we get BC = 25 . Let BD = x;Triangle ABD is similar to triangle CBA => AD/15 = x/20 and also triangle ADC is similar to triangle ACB=> AD/20 = (25-x)/15. From the 2 equations, we get x = 9 and DC = 16

We know that AREA = (semi perimeter ) * inradius

For triangle ABD, Area = 1/2 x BD X AD = 1/2 x 12 x 9 = 54 and semi perimeter = (15 + 9 + 12)/2 = 18. On using the above equation we get, inradius, r = 3.

Similarly for triangle ADC we get inradius R = 4 .

PQ = R + r = 7 cm

**Question 10: **If ABCD is a square and BCE is an equilateral triangle, what is the measure of ∠DEC?

a) 15°

b) 30°

c) 20°

d) 45°

**10) Answer (A)**

**Solution:**

According to given diagram, as triangle BCE is equilateral CE=BE=DC

Hence angle CDE = angle CED

So angle DEC = $\frac{180-150}{2} = 15$