Theorems related to triangles - class-X

Let the sides be a-d, a, a+d. Using the law of cosines and the condition that the largest angle exceeds the smallest by 90 degrees, one can derive the ratio of the sides as sqrt(7)+1 : sqrt(7) : sqrt(7)-1.

In any triangle, the distance from the orthocenter to the vertices is given by 2R cos A, 2R cos B, and 2R cos C. Summing these and relating them to the side lengths and cotangents leads to the identity PH + QH + RH = QR cot P + PR cot Q + PQ cot R.

Using the area of triangle ABC as the sum of areas of ABD and BCD, or using trigonometry in right triangles, we find AB = 6/sqrt(5).

Using Apollonius theorem and the properties of medians, the condition QR^2 + PR^2 = 5PQ^2 implies specific geometric constraints on the triangle, leading to the angle QGM being less than 45 degrees.

Given sin A = sin^2 B and 2 cos^2 A = 3 cos^2 B, substituting cos^2 A = 1 - sin^2 A = 1 - sin^4 B into the second equation allows solving for sin^2 B, which leads to A = 90 degrees.

A right-angled triangle must satisfy the Pythagorean theorem: a^2 + b^2 = c^2. For 0.5, 1.2, 1.3: 0.25 + 1.44 = 1.69, which is 1.3^2.

Calculating the area of the triangles using the determinant formula for coordinates and simplifying the ratio yields a/2.

This is a direct application of Menelaus' Theorem, which states that for a line intersecting the sides of a triangle, the product of the ratios of the segments is 1.

For option A: 7, 24, 25. Check if right triangle: 7² + 24² = 49 + 576 = 625 = 25². Also forms valid triangle (7+24 > 25). Option A is a right triangle with hypotenuse 25 cm. The question has typo ('25 cmj') but this doesn't affect the answer.

Let the sides containing the right angle be a and b, with hypotenuse c = 20 cm. The semi-perimeter is given as 24 cm, meaning the perimeter is 48 cm and a + b + c = 48, so a + b = 28 cm. Using the Pythagorean theorem, a^2 + b^2 = 20^2 = 400. From (a + b)^2 = a^2 + b^2 + 2ab, we get 28^2 = 400 + 2ab, leading to ab = 192. Solving the quadratic equations or checking options, the sides are 12 cm and 16 cm, which are not listed in options A, B, or C. Therefore, None of these is correct.

Given, $AD=3,CE=4$
Using Appaloneaus theorem for median $AD$
We have $\displaystyle{ c }^{ 2 }+{ b }^{ 2 }=2\left( \frac { { a }^{ 2 } }{ 4 } +9 \right) $   ...(1)
Using Appaloneaus theorem for median $CE$
We have $\displaystyle{ b }^{ 2 }+{ a }^{ 2 }=2\left( \frac { { c }^{ 2 } }{ 4 } +10 \right) $   ...(2)
Also, ${ a }^{ 2 }+{ c }^{ 2 }={ b }^{ 2 }$
Adding (1) and (2)
$\displaystyle 3{ b }^{ 2 }=2\left( \frac { { b }^{ 2 } }{ 4 } +25 \right) \Rightarrow { b }^{ 2 }=20$
Solving (1) and (2) we get,
$\displaystyle c=\frac { 4 }{ \sqrt { 3 }  }$ and $\displaystyle a=2\frac { 4 }{ \sqrt { 3 }  } $
Hence, area of triangle
$\displaystyle = \frac{1}{2}\left ( \frac{4}{\sqrt{3}} \right )\left ( 2\sqrt{\frac{11}{3}} \right )= \frac{4}{3}\sqrt{11}$.

The point P that minimizes the weighted sum of squared distances a(PA)^2 + b(PB)^2 + c(PC)^2 is the centroid of the triangle.

Given, $AD,BE$ and $CF$ are the medians of a $\Delta ABC$.
$\Rightarrow AB^2+AC^2=2(AD^2+BD^2)$
$\Rightarrow AB^2+AC^2=2AD^2+\displaystyle\frac{BC^2}{2}$
$\Rightarrow 2AD^2=AB^2+AC^2-\displaystyle\frac{BC^2}{2}$ -----(1)
Similarly,
$2BE^2=BC^2+BA^2-\displaystyle\frac{AC^2}{2}$ -----(2)
$2CF^2=CA^2+CB^2-\displaystyle\frac{AB^2}{2}$ -----(3)
Adding equation 1,2 and 3, we get
$2(AD^2+BE^2+CF^2)=\displaystyle\frac{3}{2}(AB^2+BC^2+CA^2)$
$\therefore (AD^2+BE^2+CF^2):(AB^2+BC^2+CA^2)=3:4$

In the right-angled triangle the sum of the other two angles is $90^o$.

Let the other side of triangle is x cm

Given ratio of sides of the triangle, $1 : \sqrt{2} : 1$
Let the triangle be $ABC$ and sides be
$AB = x$
$BC = x $
$AC = \sqrt{2}x$

By Apollonius Theorem (states that "the sum of the squares of any two sides of any triangle equals twice the square on half the third side, together with twice the square on the median bisecting the third side".) , we have

According to the Apollonius theorem, 
$AB^{2}+AC^{2}= 2[AD^{2}+\dfrac {BC}{2}^{2}]$
$1^{2}+AC^{2}= 2[3^{2}+\dfrac{2}{2}^{2}]$
$1+AC^{2}= 2[9 + 1]$
$AC^{2}=20 -1$
$AC^{2}= 19$
$AC = \sqrt{19}$
$AC = 4.35$

According to the Apollonius theorem, 
$AB^{2}+AC^{2}= 2\left [AD^{2}+\dfrac{BC}{2}^{2}\right]$
$4^{2}+6^{2}= 2\left [AD^{2}+\dfrac{2}{2}^{2}\right]$
$16+36= 2[AD^{2} + 1]$
$2AD^{2}=52 -2$
$2AD^{2}= 50$
$AD^{2} = \dfrac{50}{2}$
$AD^{2} = 25$
$AD = 5$

According to the Apollonius theorem, 
$AB^{2}+AC^{2}= 2\left [AD^{2}+\dfrac{BC}{2}^{2}\right]$
$6^{2}+8^{2}= 2\left [AD^{2}+\dfrac{2}{2}^{2}\right]$
$36+64= 2[AD^{2} + 1]$
$2AD^{2}=100 -2$
$2AD^{2}= 98$
$AD^{2} = \dfrac{98}{2}$
$AD^{2} = 49$
$AD = 7$

According to the Apollonius theorem, 
$AB^{2}+AC^{2}= 2\left [AD^{2}+\dfrac{BC}{2}^{2}\right]$
$6^{2}+4^{2}= 2\left [AD^{2}+\dfrac{2}{2}^{2}\right]$
$36+16= 2[AD^{2} + 1]$
$2AD^{2}=52 -2$
$2AD^{2}= 50$
$AD^{2} = \dfrac{50}{2}$
$AD^{2} = 25$
$AD = 5$

In any triangle, the sum of the squares on any two sides is equal to twice the square on half the third side together with twice the square on the median which bisects the third side is called Apollinius theorem.

According to the Apollonius theorem, 
$AB^{2}+AC^{2}= 2\left[AD^{2}+\dfrac{BC}{2}^{2}\right]$
$8^{2}+6^{2}= 2\left [AD^{2}+\dfrac{2}{2}^{2}\right]$
$64+36= 2[AD^{2} + 1]$
$2AD^{2}=100 -2$
$2AD^{2}= 98$
$AD^{2} = \dfrac{98}{2}$
$AD^{2} = 49$
$AD = 7$

In any triangle, the sum of the squares on any two sides is equal to twice the square on half the third side together with twice the square on the median which bisects the third side.
Apollinius theorem formula,
$a^{2}+b^{2}= 2[m^{2}+\dfrac{c}{2}^{2}]$
When the given triangle is isosceles, than $b = a$.
So, $b^{2}=m^{2}+\dfrac{c}{4}^{2}$ is the formula used for isosceles triangle.

Given : In $\triangle ABC$, $AB=4$ cm and $AC=8$ cm.

$\triangle PQR$ is a $30^o-60^o-90^o$ triangle        ....given

$\begin{array}{l} Then,, , A{ B^{ 2 } }+A{ C^{ 2 } }=? &  \ AD, , is, , median &  \ \Rightarrow A{ B^{ 2 } }+A{ C^{ 2 } }=2\left| { \frac { { B{ C^{ 2 } } } }{ 4 }  } \right| +2A{ D^{ 2 } } & \left[ \begin{array}{l} BC=2BD \ or, , BC=DC \end{array} \right]  \ \Rightarrow A{ B^{ 2 } }+A{ C^{ 2 } }=2{ \left| { \frac { { B{ C^{  } } } }{ 2 }  } \right| ^{ 2 } }+2A{ D^{ 2 } } &  \ A{ B^{ 2 } }+A{ C^{ 2 } }=2B{ D^{ 2 } }+2A{ D^{ 2 } }=2\left( { A{ D^{ 2 } }+B{ D^{ 2 } } } \right)  &  \end{array}$