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.

Advertisements
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`
APPEARS IN
संबंधित प्रश्न
In a triangle ∠ABC, ∠A = 50°, ∠B = 60° and ∠C = 70°. Find the measures of the angles of
the triangle formed by joining the mid-points of the sides of this triangle.
ABCD is a parallelogram, E and F are the mid-points of AB and CD respectively. GH is any line intersecting AD, EF and BC at G, P and H respectively. Prove that GP = PH.
Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.
In the given figure, M is mid-point of AB and DE, whereas N is mid-point of BC and DF.
Show that: EF = AC.
In trapezium ABCD, AB is parallel to DC; P and Q are the mid-points of AD and BC respectively. BP produced meets CD produced at point E.
Prove that:
- Point P bisects BE,
- PQ is parallel to AB.
In ΔABC, D, E, F are the midpoints of BC, CA and AB respectively. Find DE, if AB = 8 cm
In parallelogram PQRS, L is mid-point of side SR and SN is drawn parallel to LQ which meets RQ produced at N and cuts side PQ at M. Prove that M is the mid-point of PQ.
Show that the quadrilateral formed by joining the mid-points of the adjacent sides of a square is also a square.
The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if, ______.
D, E and F are the mid-points of the sides BC, CA and AB, respectively of an equilateral triangle ABC. Show that ∆DEF is also an equilateral triangle.
