Advertisements
Advertisements
प्रश्न
ABC is a right angles triangle with AB = 12 cm and AC = 13 cm. A circle, with centre O, has been inscribed inside the triangle.
Calculate the value of x, the radius of the inscribed circle.
Advertisements
उत्तर
In ΔABC, ∠B = 90°
OL ⊥ AB, OM ⊥ BC and ON ⊥ AC
LBNO is a square
LB = BN = OL = OM = ON = x
∴ AL = 12 – x
∴ AL = AN = 12 – x
Since ABC is a right triangle
AC2 = AB2 + BC2
`=>` 132 = 122 + BC2
`=>` 169 = 144 + BC2
`=>` BC2 = 25
`=>` BC = 5
∴ MC = 5 – x
But CM = CN
∴ CN = 5 – x
Now, AC = AN + NC
13 = (12 – x) + (5 – x)
13 = 17 – 2x
2x = 4
x = 2 cm
संबंधित प्रश्न
Calculate the area of the shaded region, if the diameter of the semicircle is equal to 14 cm. Take `pi = 22/7`
In the given figure, RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°. Calculate : ∠RNM
In the figure, given alongside, AB || CD and O is the centre of the circle. If ∠ADC = 25°; find the angle AEB. Give reasons in support of your answer.
ABCD is a cyclic quadrilateral in which AB is parallel to DC and AB is a diameter of the circle. Given ∠BED = 65°, calculate:
- ∠DAB,
- ∠BDC.
In the given figure, AB is a diameter of the circle. Chord ED is parallel to AB and ∠EAB = 63°.
Calculate:
- ∠EBA,
- ∠BCD.
In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠PQR = 58°,
Calculate:
- ∠RPQ,
- ∠STP.
Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
In the given figure, RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°. Calculate : ∠NRM
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate : ∠DBA
Also, show that the ΔAOD is an equilateral triangle.
In the given figure, BAD = 65°, ABD = 70°, BDC = 45°.
(i) Prove that AC is a diameter of the circle.
(ii) Find ACB.
