Class 9 Maths Chapter 6 Important Questions with Solutions – Measuring Space: Perimeter and Area (Ganita Manjari)

(A) The perimeter (or circumference) of a circle is given by \(\mathrm{P}=2 \pi r\) or equivalently \(\mathrm{P}=\pi d\), where \(r\) is the radius, \(d\) is the diameter, and \(\pi\) is the constant ratio of the circumference of a circle to its diameter.

i.e., \(\pi=\frac{\text { Circumference }}{\text { Diameter }}\).

(B) The length of one complete round of a circular track is its circumference: \(\mathrm{P}=\pi d\).

Given \(d=70 \mathrm{~m}\), so \(\mathrm{P}=\left(\frac{22}{7}\right) \times 70=220 \mathrm{~m}\).

Therefore, the length of one round is 220 m.

(C) Total distance \(=4.4 \mathrm{~km}=4400 \mathrm{~m}\); length of one round \(=220 \mathrm{~m}\).

Number of rounds \(=\frac{4400}{220}=20\).

Hence, the runner completes 20 rounds. Since the answer is a whole number, 4.4 km is exactly equal to 20 complete rounds of the track.

For a semicircle, arc length \(=\pi r\).

For large semicircle (diameter \(=12 \mathrm{~cm} \Rightarrow\) radius \(=6 \mathrm{~cm}\) ):

Arc length \(=\left(\frac{22}{7}\right) \times 6=\frac{132}{7} \mathrm{~cm}\).

For small semicircle \(($ diameter \(=8 \mathrm{~cm} \Rightarrow\) radius \(=4 \mathrm{~cm})\) :

Arc length \(=\left(\frac{22}{7}\right) \times 4=\frac{88}{7} \mathrm{~cm}\).

Straight segments: total straight length \(=12-8=4 \mathrm{~cm}\).

Total perimeter \(=\frac{132}{7}+\frac{88}{7}+4=\frac{220}{7}+4=\frac{248}{7}\approx35.43 \mathrm{~cm}\).

Therefore, the perimeter of the boundary is \(\frac{248}{7}\approx35.43 \mathrm{~cm}\).

In the right-angled triangle, by Baudhāyana-Pythagoras theorem,

Hypotenuse \({ }^{2}=6^{2}+8^{2}=36+64=100\)

⇒ Hypotenuse \(=10 \mathrm{~cm}\).

For a semicircle, arc length \(=\pi r\).

On side \(6 \mathrm{~cm}: r=3 \mathrm{~cm} \rightarrow\) arc length \(=3 \pi\).

On side \(8 \mathrm{~cm}: r=4 \mathrm{~cm} \rightarrow \operatorname{arc}\) length \(=4 \pi\).

On hypotenuse \(10 \mathrm{~cm}: r=5 \mathrm{~cm} \rightarrow\) arc length \(=5 \pi\).

Total perimeter \(=3 \pi+4 \pi+5 \pi=12 \pi \approx 12 \times 3.14=37.68 \mathrm{~cm}\).

(A) Side of square \(=14 \mathrm{~cm}\).

Area of square \(=14 \times 14=196 \mathrm{~cm}^{2}\).

(B) The inscribed circle has diameter equal to the side of the square, so its radius \(r=\frac{14}{2}=7 \mathrm{~cm}\).

Area of circle \(=\pi r^{2}=\left(\frac{22}{7}\right) \times 49=154 \mathrm{~cm}^{2}\).

(C) Total shaded corner area = Area of square – Area of inscribed circle \(=196-154=42 \mathrm{~cm}^{2}\).

Let the sides be \(3 k, 5 k, 7 k\). Sum \(=15 k=300 \Rightarrow k=20\).

So sides are 60 m, 100 m and 140 m.

Triangles PCD and QCD share the same base CD.

Both triangles lie between the same pair of parallel lines: CD and AB (since \(\mathrm{AB} \| \mathrm{CD}\) in a parallelogram). So they have the same height the perpendicular distance between the lines AB and CD.

Therefore \(\operatorname{Area}(\Delta \mathrm{PCD})=\left(\frac{1}{2}\right) \times \mathrm{CD} \times h\) and Area \((\Delta \mathrm{QCD})\)

Hence, Area \((\Delta \mathrm{PCD}):\) Area \((\Delta \mathrm{QCD})=1: 1\).

The rope length (14 m) is shorter than both sides of the rectangle (20 m and 16 m), so the cow can graze a quarter-circle ( \(90^{\circ}\) sector) inside the field and cannot reach beyond either adjacent side.

Grazing area \(=\left(\frac{90^{\circ}}{360^{\circ}}\right) \times \pi \times r^{2}=\left(\frac{1}{4}\right) \times\left(\frac{22}{7}\right) \times 14^{2}\)

∴ The cow can graze over an area of \(154 \mathrm{~m}^{2}\).

Brahmagupta’s formula:

where

Here \(a=39, b=60, c=52, d=25\).

Semi-perimeter

\(s-a=49, s-b=28, s-c=36, s-d=63\).