Advertisements
Advertisements
प्रश्न
In the following figure, ΔMNO is a right-angled triangle. Its legs are 6 cm and 8 cm long. Length of perpendicular NP on the side MO is ______.
पर्याय
4.8 cm
3.6 cm
2.4 cm
1.2 cm
Advertisements
उत्तर
In the following figure, ΔMNO is a right-angled triangle. Its legs are 6 cm and 8 cm long. Length of perpendicular NP on the side MO is 4.8 cm.
Explanation:
Given, ΔMNO is a right angled triangle.
So, according to Pythagoras theorem,
MO2 = MN2 + NO2
= 62 + 82
= 36 + 64
⇒ MO2 = 100
⇒ MO = `sqrt(100)`
⇒ MO = 10 cm
∴ Area of ΔMNO = `1/2` × Base × Height
⇒ `1/2` × MN × NO = `1/2` × MO × NP
⇒ `1/2` × 6 × 8 = `1/2` × 10 × NP
⇒ NP = `24/5`
⇒ NP = 4.8 cm
APPEARS IN
संबंधित प्रश्न
In each of the following find the value of 'k', for which the points are collinear.
(7, -2), (5, 1), (3, -k)
median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A (4, - 6), B (3, - 2) and C (5, 2).
Find values of k if area of triangle is 4 sq. units and vertices are (−2, 0), (0, 4), (0, k).
Find the centroid of ΔABC whose vertices are A(-1, 0) B(5, -2) and C(8,2)
Find the area of ΔABC with A(1, -4) and midpoints of sides through A being (2, -1) and (0, -1).
Show that the following points are collinear:
A(5,1), B(1, -1) and C(11, 4)
In Figure 1, PS = 3 cm, QS = 4 cm, ∠PRQ = θ, ∠PSQ = 90°, PQ ⊥ RQ and RQ = 9 cm. Evaluate tan θ.
In ∆PQR, PR = 8 cm, QR = 4 cm and PL = 5 cm. 
Find:
(i) the area of the ∆PQR
(ii) QM.
Find the coordinates of the point Q on the x-axis which lies on the perpendicular bisector of the line segment joining the points A(–5, –2) and B(4, –2). Name the type of triangle formed by the points Q, A and B.
The area of a triangle with base 4 cm and height 6 cm is 24 cm2.
