हिंदी

PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then ar (PAS) = 30 cm2.

Advertisements
Advertisements

प्रश्न

PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then ar (PAS) = 30 cm2.

विकल्प

  • True

  • False

MCQ
सत्य या असत्य
Advertisements

उत्तर

This statement is True.

Explanation:

Given, PS = 5 cm

Radius of circle = SQ = 13 cm

In right-angled ΔSPQ,

SQ2 = PQ2 + PS2   ...[By Pythagoras theorem]

(13)2 = PQ2 + (5)2

⇒ PQ2 = 169 – 25 = 144

⇒ PQ = 12 cm  ...[Taking positive square root, because length is always positive]

Now, area of ΔAPS = `1/2` × Base × Height

= `1/2 xx PS xx PQ`

= `1/2 xx 5 xx 12`

= 30 cm

So, given statement is true, if A coincides Q.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Areas of Parallelograms & Triangles - Exercise 9.2 [पृष्ठ ८८]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 9
अध्याय 9 Areas of Parallelograms & Triangles
Exercise 9.2 | Q 2. | पृष्ठ ८८

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

In the given figure, ABCD is parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.


In the given figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that

(i) ar (PQRS) = ar (ABRS)

(ii) ar (AXS) = 1/2ar (PQRS)


In the given figure, P is a point in the interior of a parallelogram ABCD. Show that

(i) ar (APB) + ar (PCD) = 1/2ar (ABCD)

(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

[Hint: Through. P, draw a line parallel to AB]


In the following figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).


ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of DC
such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:

(1)  ar ( ADEG) = ar (GBCD)

 (2)  ar (ΔEGB) = `1/6` ar (ABCD)

 (3)  ar (ΔEFC) = `1/2` ar (ΔEBF)

 (4)  ar (ΔEBG)  = ar (ΔEFC)

 (5)ΔFind what portion of the area of parallelogram is the area of EFG.


In the below fig. ABCD and AEFD are two parallelograms. Prove that
(1) PE = FQ
(2) ar (Δ APE) : ar (ΔPFA) = ar Δ(QFD) : ar (Δ PFD)
(3) ar (ΔPEA) = ar (ΔQFD)


ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (Figure). E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD) is ______.


In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q (Figure). Prove that ar (ABCD) = ar (APQD)


If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral (Figure).

[Hint: Join BD and draw perpendicular from A on BD.]


The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line is drawn to intersect AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×