Question

the area of a triangle, whose base and the corresponding altitude are 15 cm and 14 cm, is equal to area of a right triangle whose one of the sides containing the right angle is 10.5 CM find the other side of a triangle

Answer 1

Answer:

Step-by-step explanation:

A1=A2

1/2×15×14=1/2×10.5×h

21.0=10.5h

h=21.0/10.5

h=2cm

Answer 2

Step-by-step explanation:

Answer:

Given :-

Area of a triangle whose base and height are 15 cm and 4 cm is equal to area of a right angled triangle whose one of the side containg right angle is 20 cm

To Find :-

Other side

Solution :-

As we know that

\bf \red{Area \: of \: triangle = \frac{1}{2} \times b \times h}Areaoftriangle=

2

1

×b×h

\tt \implies Area = \dfrac{1}{2} \times 15 \times 4⟹Area=

2

1

×15×4

\tt \implies \: Area = 1 \times 15 \times 2⟹Area=1×15×2

\tt \implies \: Area = 30 \: {cm}^{2} ⟹Area=30cm

2

Now,

Let's find Area of second triangle

\bf \red{Area \: of \: triangle \: = \frac{1}{2} \times b \times h}Areaoftriangle=

2

1

×b×h

\tt \implies \: Area = \dfrac{1}{2} \times 20 \times h⟹Area=

2

1

×20×h

\tt \implies \: Area = 10 \times h \: ⟹Area=10×h

\tt \implies \: Area = 10h⟹Area=10h

Now,

Let's find height

\tt \implies30 = 10h⟹30=10h

\tt \implies \: h = \frac{30}{10} ⟹h=

10

30

Hence :-

Height is 3 cm