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In a right angled triangle ABC, angle B = 90o and BD is perpendicular on hypotenuse AC, AB = a, BC = b, CA = c & BD = h, then:
Solution: Since it is a right angle triangle, therefore area of right angled triangle = ½ a*b = ½ h*c.
From above relation: ab = hc. i.e. 1/h = c/ab. Now on squaring both side. 1/h2 = c2/a2b2.
c2/a2b2 = a2+b2/a2b2 = 1/a2 +1/b2. Hence 1/h2 = 1/a2 +1/b2
The median drawn from the right angled vertex to the hypotenuse of a triangle is equals to:
Solution: Median joins the vertex to the midpoint of the opposite side. Now we know that triangle in a semi-circle with one side as diameter is right triangle. Therefore median thus drawn from the right angled vertex to the midpoint of hypotenuse is also a radius. And hence, it is equals to ½ * hypotenuse as it is also diameter.
In a right angled triangle ABC, angle C = 90o and CD is the perpendicular on Hypotenuse AB. If BC = 15 cm and AC = 20cm then CD is equal to:
Solution: By Pythagoras Theorem: a2 + b2 = c2.
C2 = 152 + 202, C = 25. Now since it is a right angle triangle, therefore area of right angled triangle = ½ a*b = ½ h*c. Hence from above relation: ab = hc. Therefore CD = (15*20)/25 = 12.
In a right angled triangle ABC, angle B = 90o if P and Q are points on the sides AB & BC respectively, Then CP2 + QA2 is equals to:
AB2 + CB2 = AC2, PB2 + BQ2 = PQ2
AB2 + QB2 = QA2, CB2 + PB2 = CP2
Hence from above equations: CP2 + QA2 = AB2 + QB2 + CB2 + PB2 = AC2 + PQ2
In a triangle the length of the median drawn from the highest angle to the greatest side is equals to half of the longest side. Then the triangle is:
Solution: we know that triangle in a semi-circle with one side as diameter is right triangle. Therefore median thus drawn from the right angled vertex to the midpoint of hypotenuse is also a radius. And hence, it is equals to ½ * hypotenuse as it is also diameter which is the greatest side. Hence answer is right triangle.
In the adjoining figure if AC = 5cm, AD = 3cm & DB = 7cm. Find the length of AE. Given that DE is parallel to BC.
Solution: Since triangle ABC is similar to triangle ADE. Therefore,
AE/AD = AC/AB = 5/3+7 = 5/10.
Now, AE = ½ * AD = ½ *3 = 1.5cm
In the figure QA & PB are perpendicular to AB. If AO = 10cm, BO = 6cm and PB = 9 cm. Find AQ.
Solution: Triangle AOQ is similar to triangle POB. Hence, AQ/PB = AO/BO. AQ = AO/BO * PB = 10/6 * 9 = 15cm
In the given figure angle BDE is equals to angle ACB. BD = DE = 6cm, AC = 18cm & AD = 3cm. Find the Value of BE?
Solution: Triangle ABC is similar to BDE. (One angle is common and another is given as equal).
BE/AB = DE/AC, BE = DE/AC * AB = DE/AC * (AD+DB) = 6/8 * 9 = 3cm
In the given figure AF, BD & CE are perpendicular to AC. Then, Which of the relation is correct?
Solution: The triangle ACF is similar to triangle BCD. Hence, CB/AC = BD/AF. Similarly the triangle ABD is similar to ACE. Hence, AB/AC = BD/CE.
Now, on adding both the relation: BD/AF + BD/CE = CB/AC + AB/AC
BD (1/AF + 1/CE) = (CB+AB)/AC,
1/AF + 1/CE = 1/BD
In the given circle O is the centre of the circle and AD, AE are two tangents. BC is also a tangent at P, then which of the relation is true?
Solution: BE = BP & CD = CP, since tangent from the same point on a circle is equal.
Therefore, AE=AD. Now, AD = AC+CD = AC+CP
AE = AB + BE = AB + BP.
Therefore AE + AD = AC + CP + AB +BP
2AE = AC + AB + CP + BP = AB +AC + BC
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