Advertisements
Advertisements
प्रश्न
The given figure represents a kite with a circular and a semicircular motifs stuck on it.
The radius of a circle is 2.5 cm and the semicircle is 2 cm. If diagonals AC and BD are
of lengths 12 cm and 8 cm respectively, find the area of the:
1) Shaded part. Give your answer correct to the nearest whole number.
2) Unshaded part
उत्तर
1) Area of the shaded part = Area of the circle + area of the semicircle
`= pi(2.5)^2 + (pi(2)^2)/2`
`= pi[6.25+2]`
`= 22/7 [8.25]`
`~~ 26 cm^2`
2) Area of kite = `"product of the diagonals"/2` = `(AC xx BD)/2 = (12xx8)/2 = 48 cm^2`
Area of the unshaded part = Area of the kite - Area of the shaded part
= 48 - 26
= `22 cm^2`
APPEARS IN
संबंधित प्रश्न
(1, 5) and (–3, –1) are the co-ordinates of vertices A and C respectively of rhombus ABCD. Find the equations of the diagonals AC and BD.
Show that A(3, 2), B(6, −2) and C(2, −5) can be the vertices of a square.
- Find the co-ordinates of its fourth vertex D, if ABCD is a square.
- Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.
A line through origin meets the line x = 3y + 2 at right angles at point X. Find the co-ordinates of X.
O(0, 0), A(3, 5) and B(−5, −3) are the vertices of triangle OAB. Find the equation of median of triangle OAB through vertex O.
O(0, 0), A(3, 5) and B(−5, −3) are the vertices of triangle OAB. Find the equation of altitude of triangle OAB through vertex B.
P(3, 4), Q(7, –2) and R(–2, –1) are the vertices of triangle PQR. Write down the equation of the median of the triangle through R.
A line AB meets the x-axis at point A and y-axis at point B. The point P(−4, −2) divides the line segment AB internally such that AP : PB = 1 : 2. Find:
- the co-ordinates of A and B.
- equation of line through P and perpendicular to AB.
Point A and B have co-ordinates (7, −3) and (1, 9) respectively. Find:
- the slope of AB.
- the equation of perpendicular bisector of the line segment AB.
- the value of ‘p’ of (−2, p) lies on it.
A line is of length 10 units and one end is at the point (2, – 3). If the abscissa of the other end be 10, prove that its ordinate must be 3 or – 9.