EXERCISE 25 B
Question 1
Find the area of each of the following triangles:
Sol :
(i)
$=\frac{1}{2} \times b \times h$
$=\frac{1}{2} \times 5 \times 12$
=5×6
=30 cm 2
(ii)
$=\frac{1}{2} \times b \times h$
$=\frac{1}{2} \times 13 \times 5$
=32.5 cm 2
(iii)
$=\frac{1}{2} \times b \times h$
$=\frac{1}{2} \times 8 \times 5$
=5×4
=20 cm 2
Question 2
The base of a triangle measures 32 cm and its altitude is 45 cm. Find the area of the triangle in square metres.
Sol :
Area of triangle$=\frac{1}{2} \times 32 \times 45$
=720 cm 2
1 m2=10000 cm2
$\begin{aligned} \frac{1}{10000} \mathrm{~m}^{2} &=1 \mathrm{~cm}^{2} \\ \frac{720}{10000} m^2 &=720 \mathrm{cm}^{2} \end{aligned}$
=0.0720 m2
Question 3
Find the height of a triangle, if its area is 24 cm2 and its base is 15 cm.
Sol :
Area of triangle=24 cm2
$\frac{1}{2} \times b \times h=24$
$\frac{1}{2} \times 15 \times h=24$
$\begin{aligned} n &=\frac{24 \times 2}{15} \\ &=\frac{48}{18} \end{aligned}$
h=3.2 cm
Question 4
The area of a right-angled triangle is 60 cm2 and its smallest side is 8cm. What is its hypotenuse ?
Sol :
Area of triangle=60 cm2
$\frac{1}{2} \times 8 \times h=60$
$h=\frac{60}{4}$
h=15 cm
$x^{2}=15^{2}+8^{2}$
$x=\sqrt{225+64}$
$=\sqrt{289}$
x=17 cm
Hypotenuse=17 cm
Question 5
Calculate the shaded area of each of the following
Sol :
(i)
Area of rectangle=l×b
=48×30=1440 cm2
Area of white triangle$=\frac{1}{2} \times 20 \times(48-16)$
=10×32=320 cm2
and Also smaller triangle$=\frac{1}{2} \times 16 \times 10$
=8×10=80 cm2
Area of shaded region=Area of rectangle-(Area of unshaded region)
=1440-(320+80)
=1440-400=1040 cm2
(ii)
Area of rectangle=l×b
=14×11=154 cm2
$=\frac{1}{2} \times 11 \times 5+\frac{1}{2} \times 11 \times 4$
$=\frac{55}{2}+11 \times 2$
$=\frac{55}{2}+22$
$=\frac{55+44}{2}=\frac{99}{2}$
=49.5 cm2
Area of shaded region=154-49.5
=104.5 cm2
Question 6
Find the area of an isosceles right triangle, if one of its right sides is 18 cm long.
Sol :
Area of isosceles Δ$=\frac{1}{2} \times 18 \times 18$ (isosceles triangle have two sides of equal length)
=18×9
=162 cm2
Question 7
The area of an isosceles triangle is 300cm2. The height of the triangle is 15cm. Find the perimeter of the triangle.
Sol :
Area of isosceles triangle=300cm2
$\frac{1}{2}\times b\times h=300$
$b\times 15=300\times 2$
$b=\frac{300\times 2}{15}$
b=40cm
Using Pythagoras theorem,
H2=152+202
H2=225+400
H2=625
H=25cm
Perimeter of triangle=25+25+40
=90cm
Question 8
The area of a triangle is 1176cm2. The ratio of the base to the corresponding altitude is 3 : 4. Find the altitude of the triangle.
Sol :
Base=3x , Height=4x where x is common factor
Area of triangle=1176
$\frac{1}{2}\times b\times h=1176$
$\frac{1}{2}\times 3x\times 4x=1176$
$\frac{12x^2}{2}=1176$
6x2=1176
$x=\frac{1176}{6}$
x=√196cm
x=14cm
Height=4x=4×14=56cm
Question 9
Find the area of quadrilateral ABCD, where BD = 22 cm, CE = 7 cm and AF = 6 cm.
Sol :
Area of quadrilateral =Area of ΔDBC+Area of ΔDBA
$=\frac{1}{2}\times 22\times 7+\frac{1}{2}\times 22\times 6$
=11×7+11×6
=77+66
=143cm2
Question 10
Find the area of the triangle in each of the following where the three sides
are given as
(i) 26 cm, 28 cm, 30 cm (ii) 48 cm, 73 cm, 55 cm
Sol :
(i)
$s=\frac{26+28+30}{2}=\frac{84}{2}$
s=42
Area of triangle$=\sqrt{s(s-a)(s-b)(s-c)}$
=√42(42-26)(42-28)(42-30)
=√42×16×14×12
=√112896
=336cm2
(ii)
$s=\frac{48+73+55}{2}=\frac{176}{2}$
s=88
Area of triangle$=\sqrt{s(s-a)(s-b)(s-c)}$
=√88(88-48)(88-73)(88-55)
=√88×40×15×33
=√1742400
=1320cm2
Question 11
Find the area of an equilateral triangle whose each side is 12 cm.
Sol :
Area of equilateral triangle$=\frac{\sqrt{3}}{4}a^2$
$=\frac{\sqrt{3}}{4}12^2$
$=\frac{\sqrt{3}}{4}12\times 12$
$=\sqrt{3}\times 3\times 12$
=36√3 cm2
thnx sir yu are great
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