Answer to Question #138137 in Geometry for Hiroshima

Question

7. If a cube has an edge equal to the diagonal of another cube, find the ratio of their volumes.
8. The area of a circle circumscribing an equilateral triangle is 254.47 cm2, what is the area of the triangle?
9. The sides of an isosceles triangle are 10 and 16 m. (base). What is the length of the equal sides of a similar triangle if
the base is 10m?
10. If the radius of a circle is r, what is the apothem of an equilateral triangle inscribed in the circle?

Answer 1

7. Let the cubes be Cube A and CubeB.

Let cube A edge be a

To determine the diagonal of cube a

Hyp²=Adj²+Opp²

Hyp²=a²+a²

Hyp²=2a²

Hyp=$​ \sqrt{2a²}$

Hyp=a$​ \sqrt{2}$

Since the diagonal of Cube A is equal to the edge of Cube B, therefore Cube B has its edge to be a$​ \sqrt{2}$

Volume of Cube A = a³

Volume of Cube B =(a$​ \sqrt{2}$ )³=2a³$​ \sqrt{2}$

Ratio of the Volume of Cube A to Volume of Cube B = a³:2a³$​ \sqrt{2}$

Ratio of their volumes = 1:2$​ \sqrt{2}$

8. Area of circle = πr² = 254.47

r² = 254.47÷π = 254.47 ÷(22/7)

r² = 254.47 × 7/22 = 80.97

r=$​ \sqrt{80.97}$ = 9cm

∅ = 2 × 60° ( angle subtended by an arc at the centre of a circle is twice that of the circumference)

=> ∅ = 120°

A = ∅ ÷ 2 = 120 ÷ 2

A = 60°

Sin A = x/r

x = rSinA

x = 9 sin60° = 7.79cm

The edge of the equilateral triangle is = 2x = 2 × 7.79 = 15.58cm

h² = 15.58² - (15.58/2)²

h² = 15.58² - 7.79²

h² = 182.05

h = $​ \sqrt{182.05}$

h = 13.49cm

Area of triangle =½bh

= ½ × 15.58 × 13.49

= 105.09cm²

9. For triangles to be similar, all their sides must be in the same ratio

(Base of triangle A)/(Base of triangle B) = (Side of triangle A)/(Side of triangle)

16/10 = 10/x

cross multiply

16x = 100

x = 6.25m

The length of the equal sides of a similar triangle with base 10m is 6.25m

10.

∅ = 2 × 60° ( angle subtended by an arc at the centre of a circle is twice that of the circumference)

A = ∅ ÷ 2 = 120 ÷ 2

A = 60°

Cos A = y/r

y = rCosA

y = r × Cos60

y = r × ½

y = r/2

Apothem of the equilateral triangle inscribed in the circle is r/2.