मराठी

Triangle Abc is Right-angled at B. Given that Ab = 9 Cm, Ac = 15 Cm and D, E Are the Mid-points of the Sides Ab and Ac Respectively, Calculate (I) the Length of Bc (Ii) the Area of δAde. - Mathematics

Advertisements
Advertisements

प्रश्न

In Fig. below, triangle ABC is right-angled at B. Given that AB = 9 cm, AC = 15 cm and D,
E are the mid-points of the sides AB and AC respectively, calculate
(i) The length of BC (ii) The area of ΔADE.

 

उत्तर

 

In right  ΔABC, ∠B = 90°

By using Pythagoras theorem

            `AC^2  = AB^2+ BC^2`

⇒       `15^2 = 9^2 +BC^2`

⇒        BC =`sqrt(15^2 - 9^2)`

⇒       BC =`sqrt(225-81)`

⇒       BC =`sqrt144`

               = 12cm 

In ΔABC

D and E are midpoints of  AB and AC

∴ DE || BC, DE = `1/2` BC       [By midpoint theorem]

AD = OB = `(AB)/ 2= 9/2`  = 4 . 5cm          [ ∵ D is the midpoint of AB]

DE = `(BC)/2 = 12/2` = 6cm        

Area of ΔADE = `1/2 xxAD xx DE `

= `1/2× 4 .5 × 6 = 13.5cm^2`

 

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 13: Quadrilaterals - Exercise 13.4 [पृष्ठ ६३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 9
पाठ 13 Quadrilaterals
Exercise 13.4 | Q 7 | पृष्ठ ६३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

ABC is a triang D is a point on AB such that AD = `1/4` AB and E is a point on AC such that AE = `1/4` AC. Prove that DE = `1/4` BC.


In below Fig, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = `1/4` AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.


Fill in the blank to make the following statement correct:

The triangle formed by joining the mid-points of the sides of a right triangle is            


The following figure shows a trapezium ABCD in which AB // DC. P is the mid-point of AD and PR // AB. Prove that:

PR = `[1]/[2]` ( AB + CD)


In triangle ABC, P is the mid-point of side BC. A line through P and parallel to CA meets AB at point Q, and a line through Q and parallel to BC meets median AP at point R.
Prove that : (i) AP = 2AR
                   (ii) BC = 4QR


In trapezium ABCD, sides AB and DC are parallel to each other. E is mid-point of AD and F is mid-point of BC.
Prove that: AB + DC = 2EF.


In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively.
Prove that:
(i) Triangles HEB and FHC are congruent;
(ii) GEHF is a parallelogram.


In parallelogram ABCD, P is the mid-point of DC. Q is a point on AC such that CQ = `(1)/(4)"AC"`. PQ produced meets BC at R. Prove that

(i) R is the mid-point of BC, and

(ii) PR = `(1)/(2)"DB"`.


Side AC of a ABC is produced to point E so that CE = `(1)/(2)"AC"`. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meets AC at point P and EF at point R respectively. Prove that: 3DF = EF


In ΔABC, D, E and F are the midpoints of AB, BC and AC.
If AE and DF intersect at G, and M and N are the midpoints of GB and GC respectively, prove that DMNF is a parallelogram.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×