- 3. In the figure, BC = 10 cm, AC = 6 cm and the area of A ABC is 15 sp cm. Find (a) the value of h (b) the value of k.
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In the figure, the sides AB,BC and CA of a triangle ABC, touch a circle at P,Q and R respectively. If PA=4 cm, BP=3 cm and AC=11 cm, then the length of BC (in cm) is:
495212
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A
11
B
10
C
14
D
15
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Correct option is B)
As given
PA=4cm
BP=3cm
∴AB=4+3=7cm
AC=11cm
As PA and RA is two tangents from same external point A
∴RA=PA=4cm
⇒RC=AC−RA
⇒RC=11−4=7cm
As RC and QC is two tangents from same external point C
∴QC=RC=7cm
As BP and BQ is two tangents from same external point B
∴BP=BQ=3cm
Then BC=QC+BQ
⇒BC=7+3=10cm
Answer:
In ∆ABC,
A(∆ABC)= 15sq.cm,
Side BC is a base of the triangle and h is a height of a triangle.
BC=10cm, AC=6cm
A(∆ABC)= 1/2 base × height
15 = 1/2 × BC × h
15 = 1/2 × 10 × h
15 = 5 × h
h = 15/5
h = 3cm
In ∆DBC, angle BDC = 90°
BC=10cm, DC=6cm and BD=k
By Pythagoras theorem,
Hypt²= side² + side²
BC²= BD² + DC²
10²= k² + 6²
100 = k² + 36
100 - 36 = k²
k² = 64
taking squareroot,
k = 8cmIn ∆ABC,
A(∆ABC)= 15sq.cm,
Side BC is a base of the triangle and h is a height of a triangle.
BC=10cm, AC=6cm
A(∆ABC)= 1/2 base × height
15 = 1/2 × BC × h
15 = 1/2 × 10 × h
15 = 5 × h
h = 15/5
h = 3cm
In ∆DBC, angle BDC = 90°
BC=10cm, DC=6cm and BD=k
By Pythagoras theorem,
Hypt²= side² + side²
BC²= BD² + DC²
10²= k² + 6²
100 = k² + 36
100 - 36 = k²
k² = 64
taking square root,
k = 8cm
Step-by-step explanation:
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