#### Geometry and Mensuration

**1)**

In this figure PS is the bisector. If PQ = 8 cm, PR = 20 cm, find QS and SR.

PS is the bisector of < Q P R

\({PQ \over {PR}}\) = \({QS \over {SR}}\)

\({8 \over {20}}\) = \({QS \over {SR}}\)

\({QS \over {SR}}\) = \({2 \over {5}}\)

Option a satisfies the ratio. Hence (a).

**2)** Find the area of a triangle whose sides measure 13 cm, 16 cm and 23 cm.

Semi perimeter s = \({a+b+c \over {2}}\)

a = 13 cm, b = 16 cm and c = 23 cm

s = \({13 + 16 + 23 \over {2}}\) = 26 cm

Area = \(√26(26-13)(26-16)(26-23)\)

= 26\(√15\) cm

^{2}Hence (b).

**3)**

The length of a rectangle is increased by triple the original value and its breadth is decreased to \({1 \over {8}}\)^{th} of the original value. What is the percentage change in the area?

Let the length and breadth of rectangle be l and b respectively.

Original area = lb

New length = l + 3l = 4l

New breadth = \({b \over {8}}\)

New area of rectangle = 4l * \({b \over {8}}\) = \({lb \over {2}}\) = 0.5lb

% change in area = \({0.5lb \over {lb}}\) * 100 = 50% decrease. Hence (d).

**4)** If the radius of circle is 10.5cm, then find the area of the sector at 60˚.

^{2}= \({60˚ \over {360˚}}{22 \over {7}}\) * 10.5 * 10 .5 = 5 7.75 cm

^{2}. Hence (c).

**5)**

Find the area of the trapezium, whose parallel sides are 16 cm and 12 cm, and distance between the parallel sides is 18 cm?

Area of trapezium ** =** \({1 \over {2}}\)** *(a + b) * h**

= \({1 \over {2}}\)** *(16 + 12) * 18 = 252 ** **cm** ^{2}**. ** Hence (b).

**6)**

A hollow iron pipe is 10.5 cm long and its external diameter is 12 cm. If the thickness of the pipe is 2 cm and iron weighs 6 g/cm^{3}, then the weight of the pipe is:

External diameter = 12 cm. Therefore the radius R = 6 cm

Thickness = 2 cm. Internal radius = 6 – 2 = 4 cm

The volume of the iron = π *(R ^{2} – r ^{2} )*height = \({22 \over {7}}\) * (36 – 16) * 10.5 = 660 cm ^{3}

Therefore, the weight of iron = 660 * 6 = 3.96 kg . Hence (c).

**7)**

Find the lateral surface area of the frustum of a cone, if the area of base is 49π cm^{2}, radius of upper circle 6 cm and slant height 8 cm.

LSA of the frustum of a cone = π * (R + r) * Slant height Area of base = 49π = πr^{2}à R = 7 cm LSA = π * (7 + 6) * 8 = 104π cm^{2}. Hence (b).

**8)**

Find the ratio between the breadth and height of the cuboid whose total surface area is 1794 cm^{2}. The length and height are 15 cm and 18 cm respectively.

TSA = 2(lb + bh + lh) = 1794 cm^{2} l = 15 cm, h = 18 cm 2((15*b) + (b*18) + (15*18)) = 1794 270 + 15b + 18b = 897 33b = 627 à b = 19 cm

**9)**

A circular grassy plot of land, 48m in diameter, has a path 4.5m wide running round it on the outside. Find the cost of gravelling the path at Rs.5 per square metre.

Area of path = π (R ^{2} – r ^{2} )

= \({22 \over {7}}\) * (28.5 ^{2} - 24 ^{2} )

= \({22 \over {7}}\) * (812.25 – 576) = 22 * 33.75 = 742.5 m ^{2}

Cost of gravelling = 742.5 * 5 = Rs.3712 . 5 Hence (d).

**10)**

A lawn is in the form of an isosceles triangle. The cost of turfing it came to Rs.2400 at Rs.5 per square metre. If the base be 60m long, find the length of equal side.

Area of isosceles triangle = \({2400 \over {5}}\) = 48 0 m ^{2}

48 0 = \({1 \over {2}}\) bh

h = \({480*2 \over {60}}\) = 16 m

Equal sides of triangle = \( \sqrt {16{}{}{}{}{}{}{}{}{}{}{}{}}+30\) Hence (a).

= \( \sqrt {256}+900\) = \(√1156\) = 34m . Hence (a).

**11)**

Area of square shaped chocolate is 1764 cm^{2}. Four chocolates of circular shape of maximum size are cut off from it. Find the ratio of the remaining chocolate to the initial one.

Area of the square = a ^{2} = 1764 cm ^{2}

a = 42 cm

2D = a => 42, r = \({42 \over {4}}\) = 10.5 cm

Area of one circle = π r ^{2} = \({22 \over {7}}\) * 10.5 * 10.5 = 346.5 cm ^{2}

Total area of all four circles = 4 * 346.5 = 1386 cm ^{2}

Remaining area = 1764 – 1386 = 378 cm ^{2}

Ratio of remaining area to the initial area = 378: 1764 = 3: 14 . Hence (d).

**12)**

A cuboidal shaped building with external dimensions 8m * 6m * 13m and with uniform thickness of 1m is to be painted. Find the area of the walls and the roof to be painted in both internal and external.

Outer surface area of cuboid shaped building = 2(bh + hl) + lb = 2((6*13) + (13*8)) + (8*6) = 412m ^{2} Thickness of wall = 1m. To find inner surface area of the building, 2m to be subtracted from all dimensions. Inner surface area of the building = 2((4*11) + (11*6)) + (4*6) = 244 m ^{2} Total painting area = Outer surface area + inner surface area = 412 + 244 = 656m ^{2} Hence (c).

**13)**

A square and an equilateral triangle have the same perimeter. If diagonal of the square is 18\(√2\) cm, then area of the triangle is:

Let the side of the square be a.

Diagonal of square = \(√a\) + \(a\) = 18 \(√2\)

a = 18 cm

Perimeter of square = 4a = 72 cm = Perimeter of equilateral triangle

PQ^{2} = PS^{2} – QS^{2}PQ^{2} = 576 – 144 = 432à PQ = 12\(√3\)

Area of the triangle = \({1 \over {2}}\) * 24 *12\(√3\) = 144\(√3\)cm^{2}Hence (a).

**14)** A sphere of radius 4 cm is dropped into a cylindrical container partly filled with liquid. The radius of the vessel is 8 cm. If the sphere is submerged completely, then the surface of the liquid is raised by

_{1}

Volume of cylinder = π r

^{2}h = \({22 \over {7}}\) * 64 * h

\({22 \over {7}}\) * 64 * (h

_{1}– h) = \({4 \over {3}}\) * \({22 \over {7}}\) * 64

h

_{1}- h= \({4 \over {3}}\) * \({22 \over {7}}\) * 64 * \({7 \over {22}}\) * \({1 \over {64}}\) = 1.3cm Hence (b).

**15)**

The volume of a right circular cone varies as square of the radius of the base when the height is constant, and as the height when the base is constant. If the radius of the base is 10 cm and the height 12 cm, the volume is 720 cm^{3}, find the height of a cone whose volume is 160 cm^{3} and which stands on a base whose radius is 4cm?

Let height and radius be h and r respectively. Let V volume of CIRCULAR CONE V = xr^{2}h , where x is constant. 720 = x * 10^{2} * 12 X = \({5 \over {3}}\) V = \({5 \over {3}}\)r^{2}h For given V = 160, r = 4 160 = \({5 \over {3}}\)* 16 * h à h = 6 cm. Hence (d).