Geometry Questions with Answer - General Aptitude for GATE - Mechanical

Question 1: Rhombus of side 6 cm has an angle equal to the external angle of a regular octagon. Find the area of the rhombus. 

A. 18√2 cm²

B. 9√2 cm²

C. 15√2 cm²

D. 12√2 cm²

Ans: 18√2 cm²

Sol: Exterior angle of a regular octagon = 360° / 8 = 45°.
So the given angle of the rhombus a = 45°.
Area of a rhombus with side s and one angle a is s² sin a.
Here s = 6, so area = 6² × sin 45° = 36 × (1/√2) = 36/√2 = 18√2 cm².

Question 2: A circle inscribed in a square of side 2 has an equilateral triangle inscribed inside it. What is the ratio of areas of the equilateral triangle to that of the square? 

A. 9√3 : 16
B. 3√3 : 4
C. 9√3 : 4
D. 3√3 : 16

Ans: 3√3 : 16
Sol: The square has side 2, so the inscribed circle has radius R = 1.
For an equilateral triangle of side a, the circumradius R = a / √3.
Thus a = R√3 = 1 × √3 = √3.
Area of the equilateral triangle = (√3/4) a² = (√3/4) × (√3)² = (√3/4) × 3 = 3√3 / 4.
Area of the square = 2² = 4.
Therefore the ratio (triangle : square) = (3√3 / 4) : 4 = 3√3 : 16.

Question 3: An acute-angled isosceles triangle has two of its sides equal to 10 and 16. Find the area of this triangle. 

A. √231 units

B. 12√66 units

C. 24 units

D. 5√231 units

Ans: 5√231 units

Sol: For an acute-angled isosceles triangle, the equal sides must be such that the triangle is possible and acute.
If the equal sides were both 10 with the third side 16, then 10² + 10² = 200 < 16² = 256, which would make the triangle obtuse or impossible for the required condition.
Thus the isosceles triangle has the two equal sides 16 and 16, and the base 10.
Semiperimeter S = (16 + 16 + 10) / 2 = 21.
By Heron's formula, area = √[S(S - a)(S - b)(S - c)] = √[21 × 5 × 5 × 11] = 5√231 units.

Question 4: Three equal circles are placed inside an equilateral triangle such that any circle is tangential to two sides of the equilateral triangle and to two other circles. What is the ratio of the areas of one circle to that of the triangle 

A. π : (6+4√3)

B. 3π : (6+4√3)

C. 2π : (6+4√3)

D. π : (6+2√3)

Ans: π : (6+4√3)

Sol: Let the common radius of each small circle be r.
Each small circle has area πr².
From the configuration, the centres of the three circles form an equilateral triangle of side 2r, and each centre is at distance r from two sides of the large equilateral triangle. Carrying out the geometric relations between these distances and the altitude of the large triangle (standard construction for three mutually tangent equal circles inside an equilateral triangle) gives the area of the large triangle as (6 + 4√3) r².
Hence the ratio (area of one circle) : (area of the equilateral triangle) = πr² : (6 + 4√3) r² = π : (6 + 4√3).

Question 5: There is an equilateral triangle with a square inscribed inside it. One of the sides of the square lies on a side of the equilateral △. What is the ratio of the area of the square to that of the equilateral triangle? 

A. √3 : (5 + 4√3)

B. 2√3 : (7 + 4√3)

C. 4√3 : (7 + 4√3)

D. 4√3 : (5 + 2√3)

Ans: 4√3 : (7 + 4√3)

Sol: Let the side of the inscribed square be a and the side of the equilateral triangle be s. The altitude of the triangle is h = (√3/2) s.
At height a above the base, the width of the triangle equals the top side of the square, which is a. By linear similarity of horizontal cross-sections, a = s × (1 - a/h).
Substitute h = (√3/2) s to obtain:
a = s × [1 - (2a)/(√3 s)] ⇒ a + (2a)/√3 = s ⇒ s = a (1 + 2/√3) = a (√3 + 2)/√3.
Area of triangle = (√3/4) s² = (√3/4) a² × ((√3 + 2)² / 3).
Simplify: area(triangle) = a² × [√3(7 + 4√3) / 12].
Area(square) = a².
Therefore ratio (square : triangle) = a² : a² [√3(7 + 4√3) / 12] = 12 : [√3(7 + 4√3)] = 4√3 : (7 + 4√3).

Question 6: Consider Square S inscribed in circle C, what is the ratio of the areas of S and Q? And, Consider Circle C inscribed in Square S, what is the ratio of the areas of S and Q? 

A. 2:π, 4:π

B. 4:π, 2:π

C. 1:π, 4:π

D. 2:π, 1:π

Ans: 2:π, 4:π

Sol: Case 1 - Square inside circle (square is inscribed in the circle):
If the square has side a, its diagonal = a√2 is the diameter of the circumscribed circle, so the circle radius r = a√2 / 2 and circle area = π r² = π (a²/2).
Square area = a².
Ratio (square : circle) = a² : [π (a²/2)] = 2 : π.

Case 2 - Circle inside square (circle is inscribed in the square):
If the square has side a, the inscribed circle has diameter a so radius r = a/2 and area = π (a²/4).
Square area = a².
Ratio (square : circle) = a² : [π (a²/4)] = 4 : π.

If square is inside the circle, ratio = 2 : π. If circle is inside the square, ratio = 4 : π.

Question 7: Consider equilateral triangle T inscribed in circle C, what is ratio of the areas of T and C? Consider Circle C inscribed in equilateral triangle T, what is ratio of the areas of T and C? 

A. 3√3:π , 3√3:16π

B. 3√3:4π , 3√3:π

C. √3:π , 3√3:4π

D. √3:π , √3:16π

Ans: 3√3:4π , 3√3:π

Sol: For an equilateral triangle of side a, area(T) = (√3/4) a².
Circumradius R = a / √3 and inradius r = a / (2√3).
When the triangle T is inscribed in circle C (triangle inside circle), circle area = π R² = π (a²/3).
Ratio (T : C) = (√3/4 a²) : (π a²/3) = (√3/4) : (π/3) = 3√3 : 4π.
When the circle C is inscribed in triangle T (circle inside triangle), circle area = π r² = π (a²/12).
Ratio (T : C) = (√3/4 a²) : (π a²/12) = (√3/4) : (π/12) = 3√3 : π.

Question 8: Consider Regular Hexagon H inscribed in circle C, what is ratio of the areas of H and C? Consider Circle C inscribed in Regular Hexagon H, what is ratio of the areas of H and C? 

A. 2√3 : 3π , 3√3 : 4π

B. 3√3 : π , 3√3 : 4π

C. 3√3 : 2π, 2√3 : π

D. √3 : π , √3 : 4π

Ans: 3√3 : 2π, 2√3 : π

Sol: For a regular hexagon of side a, area(H) = (3√3/2) a².
Circumradius of the hexagon = a and inradius = (√3/2) a.
When the hexagon H is inscribed in circle C (hexagon inside circle), circle area = π a² and
Ratio (H : C) = (3√3/2 a²) : (π a²) = 3√3 : 2π.
When the circle C is inscribed in the hexagon H (circle inside hexagon), circle radius = (√3/2) a so circle area = π (3/4 a²) = 3π a² / 4 and
Ratio (H : C) = (3√3/2 a²) : (3π/4 a²) = (3√3/2) : (3π/4) = 2√3 : π.

Question 9: What is the distance between the orthocentre and the circumcenter of a triangle who sides measure 24 cm, 26 cm and 10 cm? 

A. 13 cm

B. 12 cm

C. 7.5 cm

D. √30 cm

Ans: 13 cm

Sol: The side lengths 10, 24, 26 satisfy 10² + 24² = 100 + 576 = 676 = 26², so the triangle is right-angled with hypotenuse 26.
In a right-angled triangle the orthocentre is the vertex at the right angle and the circumcentre is the midpoint of the hypotenuse. The distance from the right-angle vertex to the midpoint of the hypotenuse equals half the hypotenuse = 26 / 2 = 13 cm.

Question 10: Two circles with centres O₁ and O₂ touch each other externally at a point R. AB is a tangent to both the circles passing through R. P'Q' is another tangent to the circles touching them at P and Q respectively and also cutting AB at S. PQ measures 6 cm and the point S is at distance of 5 cms and 4 cms from the centres of the circles. What is the area of the triangle SO₁O₂? 

A. 9 cm²

B. 3(4+√7)/2 cm²

C. 27/2 cm²

D. (3√41)/2 cm² 

Ans: 3(4+√7)/2 cm²

Sol: From the diagram, SP and SR are tangents from point S to the first circle, and SR and SQ are tangents from S to the second circle. Tangents from a point to a circle are equal, so SP = SR and SQ = SR. Given PQ = 6, and PQ = SP + SQ, therefore SP = SR = SQ = 3 cm.
SR is the altitude from S to the base O₁O₂ of triangle SO₁O₂, so altitude = 3 cm.
In right triangle O₁RS, hypotenuse O₁S = 5 cm and one leg SR = 3 cm, hence O₁R = √(5² - 3²) = √(25 - 9) = 4 cm.
In right triangle O₂RS, hypotenuse O₂S = 4 cm and SR = 3 cm, hence O₂R = √(4² - 3²) = √7 cm.
Then O₁O₂ = O₁R + O₂R = 4 + √7.
Area of triangle SO₁O₂ = (1/2) × base × height = (1/2) × O₁O₂ × SR = (1/2) × (4 + √7) × 3 = 3(4 + √7) / 2 cm².