Measuring Space: Perimeter and Area — CBSE Class 9 Maths Important Questions

Answer: (a) $30\ \text{cm}^2$30 cm².

Area $=\dfrac{1}{2}\times\text{base}\times\text{height}=\dfrac{1}{2}\times10\times6=30\ \text{cm}^2$=1/2× base× height=1/2×10×6=30 cm².

Answer: (b) $44\ \text{cm}$44 cm.

Circumference $=2\pi r=2\times\dfrac{22}{7}\times7=44\ \text{cm}$=2π r=2× 22/7×7=44 cm. (Note $154\ \text{cm}^2$154 cm² would be the area.)

Answer: (a) $4\sqrt3\ \text{cm}^2$4√3 cm².

Area $=\dfrac{\sqrt3}{4}a^2=\dfrac{\sqrt3}{4}\times4^2=\dfrac{\sqrt3}{4}\times16=4\sqrt3\ \text{cm}^2$=√3/4a²=√3/4×4²=√3/4×16=4√3 cm².

Answer: (a) $84\ \text{cm}^2$84 cm².

$s=\dfrac{13+14+15}{2}=21$s=(13+14+15)/2=21. Area $=\sqrt{21(21-13)(21-14)(21-15)}=\sqrt{21\times8\times7\times6}=\sqrt{7056}=84\ \text{cm}^2$=√(21(21-13)(21-14)(21-15))=√21×8×7×6=√7056=84 cm².

Answer: (c) A is true but R is false. $\pi$π is irrational: its decimal $3.14159\ldots$3.14159… never ends and never repeats. $\dfrac{22}{7}=3.142857\ldots$22/7=3.142857… is rational and only a convenient approximation to $\pi$π, so R is false.

In one turn the wheel travels its circumference: $\pi d=\dfrac{22}{7}\times70=220\ \text{cm}$π d=22/7×70=220 cm.

In $100$100 turns: $100\times220=22000\ \text{cm}=220\ \text{m}$100×220=22000 cm=220 m.

Area $=\pi r^2=\dfrac{22}{7}\times7^2=\dfrac{22}{7}\times49=22\times7=154\ \text{cm}^2$=π r²=22/7×7²=22/7×49=22×7=154 cm².

The angle $60^\circ$60° is $\dfrac{60}{360}=\dfrac16$60/360=1/6 of a full turn, so the arc and the sector are $\dfrac16$1/6 of the whole circle.

(i) Arc length $=\dfrac16\times2\pi r=\dfrac16\times2\times\dfrac{22}{7}\times21=\dfrac16\times132=22\ \text{cm}$=1/6×2π r=1/6×2× 22/7×21=1/6×132=22 cm.

(ii) Perimeter of the sector $=$= arc $+$+ two radii $=22+21+21=64\ \text{cm}$=22+21+21=64 cm.

(iii) Area of the sector $=\dfrac16\times\pi r^2=\dfrac16\times\dfrac{22}{7}\times441=\dfrac16\times1386=231\ \text{cm}^2$=1/6× π r²=1/6× 22/7×441=1/6×1386=231 cm².

Brahmagupta's formula: area $=\sqrt{(s-a)(s-b)(s-c)(s-d)}$=√(s-a)(s-b)(s-c)(s-d), where $s=\dfrac{a+b+c+d}{2}$s=(a+b+c+d)/2.

$s=\dfrac{7+15+20+24}{2}=33$s=(7+15+20+24)/2=33, so
$\text{area}=\sqrt{(33-7)(33-15)(33-20)(33-24)}=\sqrt{26\times18\times13\times9}=\sqrt{54756}=234\ \text{cm}^2.$area=√(33-7)(33-15)(33-20)(33-24)=√26×18×13×9=√54756=234 cm².
Check: the diagonal of $25\ \text{cm}$25 cm splits the quadrilateral into two triangles with sides $7, 24, 25$7, 24, 25 and $15, 20, 25$15, 20, 25. Both are right-angled ($7^2+24^2=625=25^2$7²+24²=625=25² and $15^2+20^2=625$15²+20²=625), as an angle in a semicircle should be. Their areas are $\tfrac12\times7\times24=84$1/2×7×24=84 and $\tfrac12\times15\times20=150$1/2×15×20=150, and $84+150=234\ \text{cm}^2$84+150=234 cm². ✓

(Heron's formula is the special case of Brahmagupta's formula with one side $d=0$d=0.)

Perimeter of square $=4\times6=24\ \text{cm}$=4×6=24 cm.

Equal perimeters, so triangle perimeter $=24\ \text{cm}\Rightarrow$=24 cm⇒ side $=\dfrac{24}{3}=8\ \text{cm}$=24/3=8 cm.

Area $=\dfrac{\sqrt3}{4}a^2=\dfrac{\sqrt3}{4}\times8^2=\dfrac{\sqrt3}{4}\times64=16\sqrt3\ \text{cm}^2\approx27.7\ \text{cm}^2$=√3/4a²=√3/4×8²=√3/4×64=16√3 cm²≈27.7 cm².

By Heron's formula: sides $a=10,\ b=10,\ c=12$a=10, b=10, c=12.

$s=\dfrac{10+10+12}{2}=16\ \text{cm}$s=(10+10+12)/2=16 cm.

Area $=\sqrt{16(16-10)(16-10)(16-12)}=\sqrt{16\times6\times6\times4}=\sqrt{2304}=48\ \text{cm}^2$=√(16(16-10)(16-10)(16-12))=√16×6×6×4=√2304=48 cm².

Verification by base and height: the height from the apex bisects the base. Half base $=6\ \text{cm}$=6 cm, so

$h=\sqrt{10^2-6^2}=\sqrt{100-36}=\sqrt{64}=8\ \text{cm}$h=√(10²-6²)=√(100-36)=√64=8 cm.

Area $=\dfrac{1}{2}\times12\times8=48\ \text{cm}^2$=1/2×12×8=48 cm². Both methods agree, so the area is $48\ \text{cm}^2$48 cm².

The semicircle is attached to the breadth $=7\ \text{m}$=7 m, so its diameter is $7\ \text{m}$7 m and radius $r=\dfrac{7}{2}=3.5\ \text{m}$r=7/2=3.5 m.

(a) Area:

Rectangle $=14\times7=98\ \text{m}^2$=14×7=98 m².

Semicircle $=\dfrac{1}{2}\pi r^2=\dfrac{1}{2}\times\dfrac{22}{7}\times3.5\times3.5=\dfrac{1}{2}\times\dfrac{22}{7}\times12.25=19.25\ \text{m}^2$=1/2π r²=1/2× 22/7×3.5×3.5=1/2× 22/7×12.25=19.25 m².

Total area $=98+19.25=117.25\ \text{m}^2$=98+19.25=117.25 m².

(b) Perimeter: it consists of the three straight rectangle sides (two lengths and one breadth) plus the curved semicircle.

Straight parts $=14+7+14=35\ \text{m}$=14+7+14=35 m.

Curved part $=\dfrac{1}{2}\times2\pi r=\pi r=\dfrac{22}{7}\times3.5=11\ \text{m}$=1/2×2π r=π r=22/7×3.5=11 m.

Perimeter $=35+11=46\ \text{m}$=35+11=46 m.

(i) Perimeter $=50+50+80=180\ \text{m}$=50+50+80=180 m.

(ii) $s=\dfrac{180}{2}=90\ \text{m}$s=180/2=90 m.

(iii) Area $=\sqrt{90(90-50)(90-50)(90-80)}=\sqrt{90\times40\times40\times10}$=√(90(90-50)(90-50)(90-80))=√90×40×40×10

$=\sqrt{1440000}=1200\ \text{m}^2$=√1440000=1200 m².

(iv) Cost $=180\times20=\text{Rs }3600$=180×20=Rs 3600.