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

Answer: (a) $30\ \text{cm}^2$30 cm^2.

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 cm^2.

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 cm^2 would be the area.)

Answer: (a) $4\sqrt3\ \text{cm}^2$43 cm^2.

Area $=\dfrac{\sqrt3}{4}a^2=\dfrac{\sqrt3}{4}\times4^2=\dfrac{\sqrt3}{4}\times16=4\sqrt3\ \text{cm}^2$=3/4a^2=3/4×4^2=3/4×16=43 cm^2.

Answer: (a) $84\ \text{cm}^2$84 cm^2.

$s=\dfrac{13+14+15}{2}=21$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 cm^2.

Answer: (a) Both A and R are true and R is the correct explanation of A. Heron's formula $\sqrt{s(s-a)(s-b)(s-c)}$√s(s-a)(s-b)(s-c) uses only $a,b,c$a,b,c, so it works precisely when the height is unknown.

Perimeter $=2(l+b)=2(12+5)=2\times17=34\ \text{cm}$=2(l+b)=2(12+5)=2×17=34 cm.

Area $=l\times b=12\times5=60\ \text{cm}^2$=l× b=12×5=60 cm^2.

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

Semi-perimeter $s=\dfrac{8+15+17}{2}=\dfrac{40}{2}=20\ \text{cm}$s=8+15+17/2=40/2=20 cm.

Area $=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{20(20-8)(20-15)(20-17)}$=√s(s-a)(s-b)(s-c)=√20(20-8)(20-15)(20-17)

$=\sqrt{20\times12\times5\times3}=\sqrt{3600}=60\ \text{cm}^2$=√20×12×5×3=√3600=60 cm^2.

Area of parallelogram $=\text{base}\times\text{height}=18\times9=162\ \text{cm}^2$=base×height=18×9=162 cm^2.

Using the adjacent side as base with height $h$h: $12\times h=162\Rightarrow h=\dfrac{162}{12}=13.5\ \text{cm}$12× h=162 h=162/12=13.5 cm.

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 $=\dfrac{\sqrt3}{4}a^2=\dfrac{\sqrt3}{4}\times8^2=\dfrac{\sqrt3}{4}\times64=16\sqrt3\ \text{cm}^2\approx27.7\ \text{cm}^2$=3/4a^2=3/4×8^2=3/4×64=163 cm^227.7 cm^2.

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 cm^2.

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=√10^2-6^2=√100-36=√64=8 cm.

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

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 m^2.

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 r^2=1/2×22/7×3.5×3.5=1/2×22/7×12.25=19.25 m^2.

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

(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 m^2.

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